Answer key · Unit 4

Reactions and equations.

Each lesson’s practice questions, their correct answers and the mistakes they spot. Answers read off the Scene use the same setting the learner sees.

Misconception catalogue

Answer key 4

Lesson 1 of 4 · About 20 minutes

Conservation of mass

What happens to atoms in a chemical reaction: they change partners and none is lost, so in a sealed vessel the mass does not change. And why a candle gets lighter as it burns.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
Atoms change partners
Practice · Choose
Question
The tank holds hydrogen and oxygen. What happens to the atoms when they react and make water?

Correct answer
- Correct They let go of their partners and join up anew, and every atom stays the same element
Setup, choices and feedback
The Scene at this step
Compound
Water
Oxygen or chlorine supplied
1 × needed
Run the reaction
Elements apart
Choices
Shown in a random order.
- Correct They let go of their partners and join up anew, and every atom stays the same element
- Wrong Some are destroyed, so the water weighs less than the gases
  Misconception: Thinks mass, or atoms, can be used up or destroyed in a chemical reaction.
  Feedback from the catalogue
  In a chemical reaction no atom is destroyed and none is made: the atoms only change partners. So in a sealed vessel the products weigh exactly what the reactants weighed. When mass seems to vanish, as when a candle burns, it has left in the gases the reaction made.
- Wrong The hydrogen and oxygen atoms turn into water atoms
  Misconception: Thinks atoms turn into atoms of another element during a chemical reaction.
  Feedback from the catalogue
  A chemical reaction changes the bonds between atoms, not the atoms themselves. A hydrogen atom stays a hydrogen atom and an oxygen atom stays an oxygen atom: they only join up differently, as in the Scene. An atom becomes another element only in a nuclear reaction, when its nucleus changes.
Feedback for a right answer
The atoms only change partners. At the start each atom is bonded to its own kind, $\mathrm{H_2}$ ⁠ and $\mathrm{O_2}$ ⁠; at the end each oxygen atom is bonded to two hydrogen atoms. There is no “water atom”: water is a molecule, hydrogen and oxygen atoms joined.

Hint
Move Run the reaction slowly and follow one oxygen atom, a red one.

Worked solution
1. Before the reaction: $\mathrm{H_2}$ ⁠ molecules (two hydrogen atoms) and $\mathrm{O_2}$ ⁠ molecules (two oxygen atoms).
2. In the reaction the $\mathrm{H-H}$ ⁠ and $\mathrm{O=O}$ ⁠ bonds break, and $\mathrm{O-H}$ ⁠ bonds form.
3. After the reaction: $\mathrm{H_2O}$ ⁠ molecules. Exactly the same atoms, arranged differently.
An atom cannot vanish or become another element in a chemical reaction: that happens only when the nucleus changes, in a nuclear reaction.
Step 2 of 5
Open this step
Counting atoms
Practice · Calculate
Question
The tank started with $16$ ⁠ $\mathrm{H_2}$ ⁠ molecules and $8$ ⁠ $\mathrm{O_2}$ ⁠ molecules, and the reaction is over: the readings beside the Scene show $16$ ⁠ $\mathrm{H_2O}$ ⁠ molecules. How many hydrogen atoms are in the tank now?

Correct answer
32 ± 0.1
Setup, choices and feedback
The Scene at this step
Compound
Waterfixed
Oxygen or chlorine supplied
1 × neededfixed
Run the reaction
Reaction overfixed
Accepted within
± 0.1
Computed from the Scene at this step’s setting
Atoms of the first element in the tank, before and after/nodes/tank/outputs/atomsFirst
Wrong answers it recognises
- 16 ± 0.1
  Misconception: Ignores the small number after a symbol in a formula, counting H₂O as one H and one O.
  Feedback from the catalogue
  The small number after a symbol in a formula says how many atoms of that element are in each molecule: $\mathrm{H_2O}$ ⁠ has two hydrogen atoms and one oxygen atom. When you count atoms or work out a mass, multiply each symbol by the number after it.
Feedback for a right answer
$16 \times 2 = 32$ ⁠ hydrogen atoms, exactly as at the start: $16$ ⁠ $\mathrm{H_2}$ ⁠ molecules with two in each. The atoms moved from hydrogen molecules into water molecules, and none was lost.

Hint
Each water molecule holds two hydrogen atoms.

Worked solution
1. Each $\mathrm{H_2O}$ ⁠ molecule has $2$ ⁠ hydrogen atoms.
2. There are $16$ ⁠ water molecules: $16 \times 2 = 32$ ⁠.
3. Check: at the start there were $16$ ⁠ $\mathrm{H_2}$ ⁠ molecules, that is $16 \times 2 = 32$ ⁠ hydrogen atoms. The same number.
If you answered $16$ ⁠, you counted molecules, not atoms: the $2$ ⁠ after the H says each molecule holds two.
Step 3 of 5
Open this step
Weighing the tank
Practice · Predict, then look
Question
The sealed tank holds $16$ ⁠ hydrogen molecules and $16$ ⁠ oxygen molecules, and its contents weigh 544.2 u (atomic mass units). The gases will make liquid water, and some oxygen will be left over. How much will the tank’s contents weigh when the reaction is over?

Correct answer
- Correct Exactly 544.2 u
Setup, choices and feedback
The Scene at this step
Compound
Waterfixed
Oxygen or chlorine supplied
2 × neededfixed
Run the reaction
Elements apartfixed
Choices
- Correct Exactly 544.2 u
- Wrong Less: some of the matter is used up in the reaction
  Misconception: Thinks mass, or atoms, can be used up or destroyed in a chemical reaction.
  Feedback from the catalogue
  In a chemical reaction no atom is destroyed and none is made: the atoms only change partners. So in a sealed vessel the products weigh exactly what the reactants weighed. When mass seems to vanish, as when a candle burns, it has left in the gases the reaction made.
- Wrong More: a liquid is heavier than a gas
  Misconception: Treats a gas as having no mass, so a gas taken in or given off does not count.
  Feedback from the catalogue
  A gas has mass, like a solid or a liquid: every one of its molecules weighs something. A litre of air weighs about 1.2 g. When a metal burns it gets heavier, because it binds oxygen or chlorine from the gas, and when a candle burns it gets lighter, because carbon dioxide and water vapour leave into the air.
Once the prediction is locked in, the Scene changes
Run the reaction
Reaction overfixed
The Scene then shows: 544.2 · Mass of everything in the tank at this Reveal position, in u (the same at every position)/nodes/tank/outputs/mass
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The mass stays 544.2 u. The same $32$ ⁠ hydrogen atoms and $32$ ⁠ oxygen atoms are in the tank before and after, only joined differently: $16$ ⁠ water molecules and $8$ ⁠ oxygen molecules left over. That is the law of conservation of mass.

Worked solution
1. Before: $32$ ⁠ H atoms and $32$ ⁠ O atoms.
2. After: $16$ ⁠ $\mathrm{H_2O}$ ⁠ molecules ( $32$ ⁠ H and $16$ ⁠ O) plus $8$ ⁠ $\mathrm{O_2}$ ⁠ molecules ( $16$ ⁠ O). Again $32$ ⁠ H and $32$ ⁠ O.
3. The mass is the sum of the atoms’ masses: $32 \times 1.008 + 32 \times 15.999 = 544.2$ ⁠.
The state of matter does not change the mass. If you predicted it would change, think about the atoms: as long as the vessel is sealed, none gets in or out.
Step 4 of 5
Open this step
Open or sealed
Practice · Choose
Question
A burning candle stands on an open balance, and the reading keeps falling. Why?

Correct answer
- Correct The wax reacts with oxygen from the air, and the products, carbon dioxide and water vapour, leave into the air
Setup, choices and feedback
The Scene at this step
Compound
Water
Oxygen or chlorine supplied
1 × needed
Run the reaction
Reaction over
Choices
Shown in a random order.
- Correct The wax reacts with oxygen from the air, and the products, carbon dioxide and water vapour, leave into the air
- Wrong Burning destroys matter and turns it into heat and light
  Misconception: Thinks mass, or atoms, can be used up or destroyed in a chemical reaction.
  Feedback from the catalogue
  In a chemical reaction no atom is destroyed and none is made: the atoms only change partners. So in a sealed vessel the products weigh exactly what the reactants weighed. When mass seems to vanish, as when a candle burns, it has left in the gases the reaction made.
- Wrong The wax turns into gases, and gases weigh nothing
  Misconception: Treats a gas as having no mass, so a gas taken in or given off does not count.
  Feedback from the catalogue
  A gas has mass, like a solid or a liquid: every one of its molecules weighs something. A litre of air weighs about 1.2 g. When a metal burns it gets heavier, because it binds oxygen or chlorine from the gas, and when a candle burns it gets lighter, because carbon dioxide and water vapour leave into the air.
Feedback for a right answer
The balance weighs only what is on it. The oxygen comes from the air and the products go back into it, so the candle gets lighter. In a sealed vessel, like the tank in the Scene, the mass would not change: the products weigh exactly what the wax plus the oxygen it reacted with weighed.

Hint
Where does the oxygen come from, and where do the products go?

Worked solution
1. In burning the wax reacts with oxygen: $\text{wax} + \mathrm{O_2} \rightarrow \mathrm{CO_2} + \mathrm{H_2O}$ ⁠.
2. The oxygen was never on the balance, and the products are gases that leave it.
3. What stays on the balance is the wax not yet burned, so the reading falls.
4. Seal the candle in a vessel with its air and the reading stays put.
Heat and light are not “made of” destroyed matter. And gases have mass: the carbon dioxide and water formed weigh more than the wax that burned, because the oxygen joined them.
Step 5 of 5
Open this step
A metal that gains weight
Practice · Calculate
Question
Now the tank holds $8$ ⁠ magnesium atoms and $8$ ⁠ $\mathrm{Cl_2}$ ⁠ molecules, and they have reacted completely to $8$ ⁠ units of $\mathrm{MgCl_2}$ ⁠. A Mg atom weighs 24.3 u and a Cl atom 35.45 u. How much does all the salt formed weigh, in u?

Correct answer
761.6 ± 1
Setup, choices and feedback
The Scene at this step
Compound
Magnesium chloridefixed
Oxygen or chlorine supplied
1 × neededfixed
Run the reaction
Reaction overfixed
Accepted within
± 1
Computed from the Scene at this step’s setting
Mass of all the compound made once the reaction has run, in u/nodes/tank/outputs/productMass
Wrong answers it recognises
- 194.4 ± 1
  Misconception: Treats a gas as having no mass, so a gas taken in or given off does not count.
  Feedback from the catalogue
  A gas has mass, like a solid or a liquid: every one of its molecules weighs something. A litre of air weighs about 1.2 g. When a metal burns it gets heavier, because it binds oxygen or chlorine from the gas, and when a candle burns it gets lighter, because carbon dioxide and water vapour leave into the air.
- 478 ± 1
  Misconception: Ignores the small number after a symbol in a formula, counting H₂O as one H and one O.
  Feedback from the catalogue
  The small number after a symbol in a formula says how many atoms of that element are in each molecule: $\mathrm{H_2O}$ ⁠ has two hydrogen atoms and one oxygen atom. When you count atoms or work out a mass, multiply each symbol by the number after it.
Feedback for a right answer
A unit of $\mathrm{MgCl_2}$ ⁠ weighs $24.3 + 2 \times 35.45 = 95.2$ ⁠, and eight of them 761.6 u. That is exactly the $8$ ⁠ magnesium atoms ( $194.4$ ⁠) plus the $8$ ⁠ chlorine molecules ( $567.2$ ⁠): the metal gained all the mass of the chlorine it took up.

Hint
The salt is built from every atom that was in the tank: the metal and the gas.

Worked solution
1. One unit: $\mathrm{MgCl_2} = 24.3 + 2 \times 35.45 = 95.2\ \mathrm{u}$ ⁠.
2. Eight units: $8 \times 95.2 = 761.6\ \mathrm{u}$ ⁠.
3. Check by conservation of mass: $8 \times 24.3 + 8 \times 2 \times 35.45 = 194.4 + 567.2 = 761.6$ ⁠.
If you got $194.4$ ⁠, you weighed only the magnesium, as if the chlorine, a gas, had no mass. If you got $478$ ⁠, you counted one chlorine per unit instead of two.

Lesson 2 of 4 · About 25 minutes

Balancing equations

How to write a chemical reaction so that every atom is counted: coefficients in front of formulas, formulas you never change, elements that come in pairs, and a check by counting atoms.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
Coefficients
Practice · Choose
Question
The equation $\mathrm{H_2 + O_2 \rightarrow H_2O}$ ⁠ is not balanced: two oxygen atoms on the left, one on the right. Which equation balances it correctly?

Correct answer
- Correct $\mathrm{2H_2 + O_2 \rightarrow 2H_2O}$ ⁠
Setup, choices and feedback
The Scene at this step
Compound
Water
Oxygen or chlorine supplied
1 × needed
Run the reaction
Elements apart
Choices
Shown in a random order.
- Correct $\mathrm{2H_2 + O_2 \rightarrow 2H_2O}$ ⁠
- Wrong $\mathrm{H_2 + O_2 \rightarrow H_2O_2}$ ⁠
  Misconception: Balances an equation by changing the small numbers inside a formula instead of the coefficients in front.
  Feedback from the catalogue
  You may never change a substance’s formula: $\mathrm{H_2O_2}$ ⁠ is hydrogen peroxide, a different substance from water. You balance only with coefficients, the big numbers in front of the formulas, which say how many molecules take part: $\mathrm{2H_2 + O_2 \rightarrow 2H_2O}$ ⁠.
- Wrong $\mathrm{H_2 + O \rightarrow H_2O}$ ⁠
  Misconception: Writes hydrogen, oxygen or chlorine as single atoms (H, O, Cl) instead of the two-atom molecules they are.
  Feedback from the catalogue
  Hydrogen, oxygen, nitrogen and the halogens such as chlorine do not exist as lone atoms: each is a gas of two-atom molecules, $\mathrm{H_2}$ ⁠, $\mathrm{O_2}$ ⁠, $\mathrm{Cl_2}$ ⁠. In the Scene they come in pairs. In an equation you write them that way, and balance with the coefficients.
- Wrong $\mathrm{2H_2 + O_2 \rightarrow H_2O}$ ⁠
  Feedback
  Count the hydrogen: $2 \times 2 = 4$ ⁠ atoms on the left, only $2$ ⁠ on the right. The water needs doubling too.
Feedback for a right answer
$\mathrm{2H_2 + O_2 \rightarrow 2H_2O}$ ⁠: $4$ ⁠ hydrogen atoms and $2$ ⁠ oxygen atoms on each side. The readings beside the Scene show the same equation, and the tank acts it out: $16$ ⁠ hydrogen molecules and $8$ ⁠ oxygen make $16$ ⁠ water molecules.

Hint
You may change only the numbers in front of the formulas.

Worked solution
1. Oxygen: $2$ ⁠ on the left, $1$ ⁠ on the right. Double the water: $\mathrm{H_2 + O_2 \rightarrow 2H_2O}$ ⁠.
2. Now hydrogen: $2 \times 2 = 4$ ⁠ on the right, $2$ ⁠ on the left. Double the hydrogen: $\mathrm{2H_2 + O_2 \rightarrow 2H_2O}$ ⁠.
3. Check: H $4$ ⁠ left, $4$ ⁠ right; O $2$ ⁠ left, $2$ ⁠ right.
$\mathrm{H_2O_2}$ ⁠ is hydrogen peroxide, another substance: you never change the formula itself. And oxygen does not come as a lone atom, O, but as the molecule $\mathrm{O_2}$ ⁠.
Step 2 of 5
Open this step
Two from one
Practice · Calculate
Question
The tank holds $8$ ⁠ $\mathrm{H_2}$ ⁠ molecules and $8$ ⁠ $\mathrm{Cl_2}$ ⁠ molecules, exactly enough. Each $\mathrm{HCl}$ ⁠ molecule holds one hydrogen atom and one chlorine atom. How many $\mathrm{HCl}$ ⁠ molecules will form?

Correct answer
16 ± 0.1
Setup, choices and feedback
The Scene at this step
Compound
Hydrogen chloridefixed
Oxygen or chlorine supplied
1 × neededfixed
Run the reaction
Elements apartfixed
Accepted within
± 0.1
Computed from the Scene at this step’s setting
Product units made once the reaction has run/nodes/tank/outputs/formed
Wrong answers it recognises
- 8 ± 0.1
  Misconception: Assumes reactants always combine one to one, ignoring the coefficients of the balanced equation.
  Feedback from the catalogue
  Reactants do not always combine one to one. The coefficients of the balanced equation give the ratio: in $\mathrm{2H_2 + O_2 \rightarrow 2H_2O}$ ⁠ each oxygen molecule takes two hydrogen molecules, and each hydrogen molecule gives one water molecule. Use that ratio to see how much reacts and what forms.
Feedback for a right answer
$16$ ⁠: there are $16$ ⁠ hydrogen atoms and $16$ ⁠ chlorine atoms, and each $\mathrm{HCl}$ ⁠ takes one of each. So the balanced equation is $\mathrm{H_2 + Cl_2 \rightarrow 2HCl}$ ⁠: each hydrogen molecule and chlorine molecule give two $\mathrm{HCl}$ ⁠ molecules.

Hint
Count the hydrogen atoms: each one ends up in a different $\mathrm{HCl}$ ⁠ molecule.

Worked solution
1. Hydrogen atoms: $8 \times 2 = 16$ ⁠. Chlorine atoms: $8 \times 2 = 16$ ⁠.
2. Each $\mathrm{HCl}$ ⁠ has one H and one Cl, so $16$ ⁠ molecules form.
3. Each $\mathrm{H_2}$ ⁠ gives two: $\mathrm{H_2 + Cl_2 \rightarrow 2HCl}$ ⁠.
If you answered $8$ ⁠, you assumed one molecule of each reactant gives one molecule of product. The coefficient $2$ ⁠ says otherwise.
Step 3 of 5
Open this step
Just enough chlorine
Practice · Reach the target
Question
The tank holds $8$ ⁠ sodium atoms, and they react by $\mathrm{2Na + Cl_2 \rightarrow 2NaCl}$ ⁠. Right now there is twice the chlorine needed, and some is left over. Change Oxygen or chlorine supplied until nothing but salt is left when the reaction ends, then press Check.

Goal
- 0 ± 0.1
  At the start of the step: 0
- 0 ± 0.1
  At the start of the step: 4
Setup, choices and feedback
The Scene at this step
Compound
Table saltfixed
Oxygen or chlorine supplied
2 × needed
Run the reaction
Reaction overfixed
Read from the Scene at
Units of the first element left unreacted (molecules of a gas, atoms of a metal)/nodes/tank/outputs/leftoverFirst
Read from the Scene at
Molecules of the second element left unreacted/nodes/tank/outputs/leftoverSecond
One setting that reaches it
Oxygen or chlorine supplied
1 × needed
Feedback for a right answer
Exactly $4$ ⁠ $\mathrm{Cl_2}$ ⁠ molecules, once what is needed: each molecule gives two chlorine atoms, one per sodium atom. Eight units of $\mathrm{NaCl}$ ⁠ formed and nothing is left over.

Feedback for a wrong answer
Not yet. At the end there is still sodium or chlorine left. The coefficients say: one chlorine molecule for every two sodium atoms.

Hint
By the equation, how many $\mathrm{Cl_2}$ ⁠ molecules does each pair of sodium atoms need?

Worked solution
1. By the equation: $2$ ⁠ Na atoms per $\mathrm{Cl_2}$ ⁠ molecule.
2. Eight Na atoms need $8 \div 2 = 4$ ⁠ $\mathrm{Cl_2}$ ⁠ molecules.
3. That is exactly 1 × what is needed: set Oxygen or chlorine supplied to 1.
Any less and sodium is left over; any more and chlorine is.
Step 4 of 5
Open this step
Formulas set by charges
Practice · Choose
Question
Magnesium gives up two electrons to form the ion $\mathrm{Mg^{2+}}$ ⁠, and chlorine takes one to form $\mathrm{Cl^-}$ ⁠. What is the balanced equation for magnesium and chlorine?

Correct answer
- Correct $\mathrm{Mg + Cl_2 \rightarrow MgCl_2}$ ⁠
Setup, choices and feedback
The Scene at this step
Compound
Magnesium chloride
Oxygen or chlorine supplied
1 × needed
Run the reaction
Elements apart
Choices
Shown in a random order.
- Correct $\mathrm{Mg + Cl_2 \rightarrow MgCl_2}$ ⁠
- Wrong $\mathrm{2Mg + Cl_2 \rightarrow 2MgCl}$ ⁠
  Misconception: Balances an equation by changing the small numbers inside a formula instead of the coefficients in front.
  Feedback from the catalogue
  You may never change a substance’s formula: $\mathrm{H_2O_2}$ ⁠ is hydrogen peroxide, a different substance from water. You balance only with coefficients, the big numbers in front of the formulas, which say how many molecules take part: $\mathrm{2H_2 + O_2 \rightarrow 2H_2O}$ ⁠.
- Wrong $\mathrm{Mg + 2Cl \rightarrow MgCl_2}$ ⁠
  Misconception: Writes hydrogen, oxygen or chlorine as single atoms (H, O, Cl) instead of the two-atom molecules they are.
  Feedback from the catalogue
  Hydrogen, oxygen, nitrogen and the halogens such as chlorine do not exist as lone atoms: each is a gas of two-atom molecules, $\mathrm{H_2}$ ⁠, $\mathrm{O_2}$ ⁠, $\mathrm{Cl_2}$ ⁠. In the Scene they come in pairs. In an equation you write them that way, and balance with the coefficients.
Feedback for a right answer
$\mathrm{Mg + Cl_2 \rightarrow MgCl_2}$ ⁠, balanced as it stands: one Mg and two Cl on each side. The formula $\mathrm{MgCl_2}$ ⁠ is fixed by the charges, $+2$ ⁠ and two $-1$ ⁠, not by what suits the balancing. In the Scene each magnesium atom gathers two chloride ions.

Hint
How many $\mathrm{Cl^-}$ ⁠ ions balance the charge of one $\mathrm{Mg^{2+}}$ ⁠?

Worked solution
1. The product’s formula: $\mathrm{Mg^{2+}}$ ⁠ and two $\mathrm{Cl^-}$ ⁠ balance the charge, so $\mathrm{MgCl_2}$ ⁠.
2. Write the reactants as they exist: magnesium as a metal, Mg, and chlorine as a gas, $\mathrm{Cl_2}$ ⁠.
3. Count: left Mg $1$ ⁠, Cl $2$ ⁠; right Mg $1$ ⁠, Cl $2$ ⁠. Balanced, every coefficient $1$ ⁠.
$\mathrm{MgCl}$ ⁠ does not exist: it would leave a charge of $+1$ ⁠. And chlorine does not come as lone atoms, but as $\mathrm{Cl_2}$ ⁠ molecules.
Step 5 of 5
Open this step
Balancing by counting
Practice · Calculate
Question
Cooking gas is mostly methane. Balance its burning: $\mathrm{CH_4 + {?}\,O_2 \rightarrow CO_2 + 2H_2O}$ ⁠. What is the coefficient of $\mathrm{O_2}$ ⁠?

Correct answer
2 ± 0.01
Setup, choices and feedback
The Scene at this step
Compound
Water
Oxygen or chlorine supplied
1 × needed
Run the reaction
Reaction over
Accepted within
± 0.01
A fixed value in the question
Wrong answers it recognises
- 4 ± 0.01
  Misconception: Writes the number of atoms needed as the coefficient of a two-atom molecule, as 4 O₂ for four oxygen atoms.
  Feedback from the catalogue
  The coefficient counts molecules, not atoms. Each $\mathrm{O_2}$ ⁠ molecule brings two oxygen atoms, so four oxygen atoms need $2\,\mathrm{O_2}$ ⁠, not $4\,\mathrm{O_2}$ ⁠. Divide the atoms you need by the atoms in each molecule.
- 1.5 ± 0.01
  Misconception: Forgets that a coefficient multiplies every atom of the formula after it, counting 2H₂O as one oxygen.
  Feedback from the catalogue
  A coefficient multiplies the whole formula after it: $\mathrm{2H_2O}$ ⁠ is two water molecules, so it holds $2 \times 2 = 4$ ⁠ hydrogen atoms and $2 \times 1 = 2$ ⁠ oxygen atoms. When you check a balance, count each element on each side: coefficient times the number after the symbol.
Feedback for a right answer
$\mathrm{CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O}$ ⁠. On the right $2 + 2 \times 1 = 4$ ⁠ oxygen atoms, and two $\mathrm{O_2}$ ⁠ molecules bring exactly four.

Hint
Count the oxygen atoms on the right: in the $\mathrm{CO_2}$ ⁠ and in both water molecules.

Worked solution
1. Carbon: $1$ ⁠ left, $1$ ⁠ right. Hydrogen: $4$ ⁠ left, $2 \times 2 = 4$ ⁠ right. Both already balance.
2. Oxygen on the right: $2$ ⁠ in $\mathrm{CO_2}$ ⁠, and $2 \times 1 = 2$ ⁠ in $\mathrm{2H_2O}$ ⁠. That makes $4$ ⁠.
3. Each $\mathrm{O_2}$ ⁠ brings $2$ ⁠ atoms, so you need $4 \div 2 = 2$ ⁠ molecules.
If you answered $4$ ⁠, you wrote the number of atoms as the coefficient of a two-atom molecule. If you answered $1.5$ ⁠, you counted only one oxygen in the two water molecules.

Lesson 3 of 4 · About 25 minutes

Reacting masses

From a balanced equation to grams: relative molecular mass, the fixed ratio by mass between reactants, and the three steps from grams to moles, through the coefficients, and back to grams.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
Relative molecular mass
Practice · Calculate
Question
The relative atomic mass of hydrogen is 1.008, and of oxygen 15.999. What is the relative molecular mass of water, $\mathrm{H_2O}$ ⁠?

Correct answer
18.02 ± 0.05
Setup, choices and feedback
The Scene at this step
Compound
Waterfixed
Oxygen or chlorine supplied
1 × needed
Run the reaction
Reaction over
Accepted within
± 0.05
Computed from the Scene at this step’s setting
Relative formula mass of one product unit, in u/nodes/tank/outputs/formulaMass
Wrong answers it recognises
- 17.01 ± 0.05
  Misconception: Ignores the small number after a symbol in a formula, counting H₂O as one H and one O.
  Feedback from the catalogue
  The small number after a symbol in a formula says how many atoms of that element are in each molecule: $\mathrm{H_2O}$ ⁠ has two hydrogen atoms and one oxygen atom. When you count atoms or work out a mass, multiply each symbol by the number after it.
Feedback for a right answer
$2 \times 1.008 + 15.999 = 18.015$ ⁠. So a mole of water weighs 18.0 g: $18$ ⁠ grams of water, a little over a tablespoon, hold 6.02 × 10²³ molecules.

Hint
Add the masses of every atom in the molecule, each symbol times the number after it.

Worked solution
1. In $\mathrm{H_2O}$ ⁠: $2$ ⁠ H atoms and $1$ ⁠ O atom.
2. $2 \times 1.008 = 2.016$ ⁠.
3. $2.016 + 15.999 = 18.015$ ⁠.
If you got $17.007$ ⁠, you counted one hydrogen atom: the $2$ ⁠ after the H belongs in the sum.
Step 2 of 5
Open this step
Gram for gram
Practice · Calculate
Question
The readings beside the Scene show how many grams of chlorine combine with one gram of magnesium. A 12 g magnesium ribbon burns in chlorine until all of it is $\mathrm{MgCl_2}$ ⁠. How many grams of chlorine reacted?

Correct answer
35.01 g ± 1%
Setup, choices and feedback
The Scene at this step
Compound
Magnesium chloridefixed
Oxygen or chlorine supplied
1 × needed
Run the reaction
Reaction over
Accepted within
± 1%
Computed from the Scene at this step’s setting
Grams of the second element per gram of the first in the compound/nodes/tank/outputs/massRatio
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 24 g ± 1%
  Misconception: Takes the ratio of coefficients, or of atoms, as the ratio of masses.
  Feedback from the catalogue
  The coefficients say how many particles (or moles) react, not how many grams. A chlorine atom weighs 1.46 times a magnesium atom, so even though the ratio is $1 : 2$ ⁠ in atoms, it is $1 : 2.92$ ⁠ by mass. To get grams, multiply each substance’s moles by its molar mass.
- 12 g ± 1%
  Misconception: Takes the ratio of coefficients, or of atoms, as the ratio of masses.
  Feedback from the catalogue
  The coefficients say how many particles (or moles) react, not how many grams. A chlorine atom weighs 1.46 times a magnesium atom, so even though the ratio is $1 : 2$ ⁠ in atoms, it is $1 : 2.92$ ⁠ by mass. To get grams, multiply each substance’s moles by its molar mass.
Feedback for a right answer
$12 \times 2.92 \approx 35.0$ ⁠, that is 35.0 g of chlorine. In moles: 12 g of magnesium is $12 \div 24.3 = 0.494$ ⁠ mol, which takes $0.494$ ⁠ mol of $\mathrm{Cl_2}$ ⁠, that is $0.494 \times 70.9 = 35.0$ ⁠ grams.

Hint
The mass ratio stays fixed: if 1 g of magnesium takes 2.92 g of chlorine, how much do 12 g take?

Worked solution
1. From the readings: a gram of magnesium combines with $2.92$ ⁠ grams of chlorine.
2. The ratio is fixed, so $12 \times 2.917 = 35.0$ ⁠ grams.
3. The same in moles: $n(\mathrm{Mg}) = \dfrac{12}{24.3} = 0.494$ ⁠; by $\mathrm{Mg + Cl_2 \rightarrow MgCl_2}$ ⁠ also $n(\mathrm{Cl_2}) = 0.494$ ⁠; $m = 0.494 \times 70.9 = 35.0$ ⁠.
If you answered $24$ ⁠ or $12$ ⁠, you turned the ratio of particles, $1 : 2$ ⁠ in atoms or $1 : 1$ ⁠ in molecules, into a ratio of masses. But a chlorine atom is heavier than a magnesium atom.
Step 3 of 5
Open this step
The product's mass
Practice · Calculate
Question
The same 12 g magnesium ribbon has all become $\mathrm{MgCl_2}$ ⁠. How many grams of salt formed?

Correct answer
47.01 g ± 1%
Setup, choices and feedback
The Scene at this step
Compound
Magnesium chloridefixed
Oxygen or chlorine supplied
1 × needed
Run the reaction
Reaction over
Accepted within
± 1%
Computed from the Scene at this step’s setting
Grams of compound made per gram of the first element/nodes/tank/outputs/compoundPerFirst
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 35.01 g ± 1%
  Misconception: Gives the mass of one reactant as the mass of the product, leaving the other reactant's mass out.
  Feedback from the catalogue
  All the atoms of both reactants go into the product, and so does the mass of both: the product weighs the magnesium plus the chlorine that reacted with it. That is conservation of mass, and it makes a check: the reactants that reacted weigh what the products weigh.
- 12 g ± 1%
  Misconception: Treats a gas as having no mass, so a gas taken in or given off does not count.
  Feedback from the catalogue
  A gas has mass, like a solid or a liquid: every one of its molecules weighs something. A litre of air weighs about 1.2 g. When a metal burns it gets heavier, because it binds oxygen or chlorine from the gas, and when a candle burns it gets lighter, because carbon dioxide and water vapour leave into the air.
Feedback for a right answer
$12 + 35.0 = 47.0$ ⁠, that is 47.0 g. The same in moles: $0.494$ ⁠ mol of $\mathrm{MgCl_2}$ ⁠ times 95.2 g/mol.

Hint
Conservation of mass: the salt weighs what went into it.

Worked solution
1. From the previous question: $12$ ⁠ grams of magnesium react with $35.0$ ⁠ grams of chlorine.
2. Conservation of mass: $m(\mathrm{MgCl_2}) = 12 + 35.0 = 47.0$ ⁠ grams.
3. Check in moles: $n = 0.494$ ⁠; $M(\mathrm{MgCl_2}) = 24.3 + 2 \times 35.45 = 95.2$ ⁠; $m = 0.494 \times 95.2 = 47.0$ ⁠.
If you answered $35.0$ ⁠, you gave only the chlorine. If you answered $12$ ⁠, only the magnesium. The salt is built from both.
Step 4 of 5
Open this step
Through moles
Practice · Calculate
Question
4.0 g of hydrogen burn in plenty of oxygen by $\mathrm{2H_2 + O_2 \rightarrow 2H_2O}$ ⁠. How many grams of water form? ( $M(\mathrm{H_2}) = 2.016$ ⁠, $M(\mathrm{H_2O}) = 18.015$ ⁠, in g/mol.)

Correct answer
35.74 g ± 1%
Setup, choices and feedback
The Scene at this step
Compound
Waterfixed
Oxygen or chlorine supplied
1 × needed
Run the reaction
Reaction over
Accepted within
± 1%
Computed from the Scene at this step’s setting
Grams of compound made per gram of the first element/nodes/tank/outputs/compoundPerFirst
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 4 g ± 1%
  Misconception: Takes the ratio of coefficients, or of atoms, as the ratio of masses.
  Feedback from the catalogue
  The coefficients say how many particles (or moles) react, not how many grams. A chlorine atom weighs 1.46 times a magnesium atom, so even though the ratio is $1 : 2$ ⁠ in atoms, it is $1 : 2.92$ ⁠ by mass. To get grams, multiply each substance’s moles by its molar mass.
Feedback for a right answer
35.7 g: $\dfrac{4.0}{2.016} = 1.98$ ⁠ mol of hydrogen give $1.98$ ⁠ mol of water, $1.98 \times 18.015 = 35.7$ ⁠ grams. Nearly nine times the hydrogen, because most of water’s mass is the oxygen.

Hint
Grams → moles → ratio of coefficients → moles → grams.

Worked solution
1. Moles of hydrogen: $n(\mathrm{H_2}) = \dfrac{4.0}{2.016} = 1.984$ ⁠.
2. The ratio in the equation: $\mathrm{2H_2 : 2H_2O}$ ⁠, that is $1 : 1$ ⁠. So $n(\mathrm{H_2O}) = 1.984$ ⁠.
3. Grams of water: $m = 1.984 \times 18.015 = 35.7$ ⁠.
If you answered $4.0$ ⁠, you took the $2 : 2$ ⁠ ratio of coefficients as a ratio of masses. The coefficients count molecules, and a water molecule is nine times heavier than a hydrogen molecule.
Step 5 of 5
Open this step
The other way round
Practice · Calculate
Question
How many grams of hydrogen does it take to make 90 g of water?

Correct answer
10.07 g ± 1.5%
Setup, choices and feedback
The Scene at this step
Compound
Water
Oxygen or chlorine supplied
1 × needed
Run the reaction
Reaction over
Accepted within
± 1.5%
A fixed value in the question
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 5 g ± 1.5%
  Misconception: Stops at the number of moles when the question asks for a mass in grams.
  Feedback from the catalogue
  The number of moles is a step on the way, not the answer. If the question asks for grams, finish by multiplying the moles by the molar mass of the substance it asks about: $m = n \times M$ ⁠.
- 90 g ± 1.5%
  Misconception: Takes the ratio of coefficients, or of atoms, as the ratio of masses.
  Feedback from the catalogue
  The coefficients say how many particles (or moles) react, not how many grams. A chlorine atom weighs 1.46 times a magnesium atom, so even though the ratio is $1 : 2$ ⁠ in atoms, it is $1 : 2.92$ ⁠ by mass. To get grams, multiply each substance’s moles by its molar mass.
Feedback for a right answer
10.1 g: $\dfrac{90}{18.015} = 5.00$ ⁠ mol of water need $5.00$ ⁠ mol of hydrogen, $5.00 \times 2.016 = 10.1$ ⁠ grams.

Hint
This time go the other way: from grams of water to moles of water, then to hydrogen.

Worked solution
1. Moles of water: $n(\mathrm{H_2O}) = \dfrac{90}{18.015} = 4.996$ ⁠.
2. By $\mathrm{2H_2 + O_2 \rightarrow 2H_2O}$ ⁠, the ratio is $1 : 1$ ⁠: $n(\mathrm{H_2}) = 4.996$ ⁠.
3. Grams of hydrogen: $m = 4.996 \times 2.016 = 10.07$ ⁠.
If you answered $5$ ⁠, you stopped at the moles. Multiply by hydrogen’s molar mass to get grams.

Lesson 4 of 4 · About 30 minutes

Limiting reagent and yield

Which runs out first and what is left in excess, why more of one reactant does not always give more product, and how to compare what you got in the lab with what you should have got.

Open the lesson

The 6 steps and their answers

Step 1 of 6
Open this step
Which runs out first
Practice · Predict, then look
Question
The tank holds $16$ ⁠ hydrogen molecules and $12$ ⁠ oxygen molecules, and they react by $\mathrm{2H_2 + O_2 \rightarrow 2H_2O}$ ⁠. Which runs out first?

Correct answer
- Correct The hydrogen
Setup, choices and feedback
The Scene at this step
Compound
Waterfixed
Oxygen or chlorine supplied
1.5 × neededfixed
Run the reaction
Elements apartfixed
Choices
- Correct The hydrogen
- Wrong The oxygen, because there is less of it
  Misconception: Thinks the reactant there is less of (fewer particles, or a smaller mass) is always the one that runs out.
  Feedback from the catalogue
  The limiting reagent is the one that runs out by the ratio in the equation, not the one there is less of. Twelve oxygen molecules need $24$ ⁠ hydrogen molecules, so $16$ ⁠ hydrogen molecules run out first, though there are more of them. Compare what you have with what you need: divide each amount by its coefficient, and the smaller result limits.
- Wrong Both run out together
  Misconception: Assumes reactants always combine one to one, ignoring the coefficients of the balanced equation.
  Feedback from the catalogue
  Reactants do not always combine one to one. The coefficients of the balanced equation give the ratio: in $\mathrm{2H_2 + O_2 \rightarrow 2H_2O}$ ⁠ each oxygen molecule takes two hydrogen molecules, and each hydrogen molecule gives one water molecule. Use that ratio to see how much reacts and what forms.
Once the prediction is locked in, the Scene changes
Run the reaction
Reaction overfixed
The Scene then shows: 4 · Molecules of the second element left unreacted/nodes/tank/outputs/leftoverSecond
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The hydrogen runs out, and $4$ ⁠ oxygen molecules are left. Each oxygen molecule needs two hydrogen molecules, and $12$ ⁠ oxygen molecules would need $24$ ⁠. There are only $16$ ⁠, so hydrogen is the limiting reagent, and oxygen is in excess.

Worked solution
1. Oxygen the hydrogen needs: $16 \div 2 = 8$ ⁠ $\mathrm{O_2}$ ⁠ molecules. There are $12$ ⁠: enough, with $4$ ⁠ to spare.
2. Hydrogen the oxygen needs: $12 \times 2 = 24$ ⁠ $\mathrm{H_2}$ ⁠ molecules. There are $16$ ⁠: not enough.
3. So hydrogen runs out first and limits the reaction. Sixteen water molecules form.
If you predicted the oxygen would run out, you looked at the smaller number. What counts is how much there is against how much is needed.
Step 2 of 6
Open this step
What is left
Practice · Calculate
Question
This time the tank holds $16$ ⁠ hydrogen molecules and $14$ ⁠ oxygen molecules. How many oxygen molecules will be left when the reaction is over?

Correct answer
6 ± 0.1
Setup, choices and feedback
The Scene at this step
Compound
Waterfixed
Oxygen or chlorine supplied
1.75 × neededfixed
Run the reaction
Elements apartfixed
Accepted within
± 0.1
Computed from the Scene at this step’s setting
Molecules of the second element left unreacted/nodes/tank/outputs/leftoverSecond
Wrong answers it recognises
- 0 ± 0.1
  Misconception: Assumes reactants always combine one to one, ignoring the coefficients of the balanced equation.
  Feedback from the catalogue
  Reactants do not always combine one to one. The coefficients of the balanced equation give the ratio: in $\mathrm{2H_2 + O_2 \rightarrow 2H_2O}$ ⁠ each oxygen molecule takes two hydrogen molecules, and each hydrogen molecule gives one water molecule. Use that ratio to see how much reacts and what forms.
- 2 ± 0.1
  Misconception: Assumes reactants always combine one to one, ignoring the coefficients of the balanced equation.
  Feedback from the catalogue
  Reactants do not always combine one to one. The coefficients of the balanced equation give the ratio: in $\mathrm{2H_2 + O_2 \rightarrow 2H_2O}$ ⁠ each oxygen molecule takes two hydrogen molecules, and each hydrogen molecule gives one water molecule. Use that ratio to see how much reacts and what forms.
Feedback for a right answer
$6$ ⁠ molecules. Sixteen hydrogen molecules use $16 \div 2 = 8$ ⁠ oxygen molecules, leaving $14 - 8 = 6$ ⁠. Add as much oxygen as you like: still only $16$ ⁠ water molecules form.

Hint
Hydrogen limits: how much oxygen does it use?

Worked solution
1. Hydrogen needs $16 \div 2 = 8$ ⁠ $\mathrm{O_2}$ ⁠ molecules, and there are $14$ ⁠: hydrogen limits.
2. Oxygen that reacts: $8$ ⁠.
3. Oxygen left over: $14 - 8 = 6$ ⁠.
If you answered $0$ ⁠ or $2$ ⁠, you assumed hydrogen and oxygen react one to one. By the equation, each $\mathrm{O_2}$ ⁠ takes two $\mathrm{H_2}$ ⁠.
Step 3 of 6
Open this step
The excess stays in excess
Practice · Predict, then look
Question
$8$ ⁠ magnesium atoms, and this time twice the chlorine needed: $16$ ⁠ $\mathrm{Cl_2}$ ⁠ molecules. What will the tank hold when the reaction is over?

Correct answer
- Correct $8$ ⁠ units of $\mathrm{MgCl_2}$ ⁠, and $8$ ⁠ chlorine molecules in excess
Setup, choices and feedback
The Scene at this step
Compound
Magnesium chloridefixed
Oxygen or chlorine supplied
2 × neededfixed
Run the reaction
Elements apartfixed
Choices
- Correct $8$ ⁠ units of $\mathrm{MgCl_2}$ ⁠, and $8$ ⁠ chlorine molecules in excess
- Wrong $16$ ⁠ units of $\mathrm{MgCl_2}$ ⁠
  Misconception: Thinks adding more of one reactant always makes more product, even after the other has run out.
  Feedback from the catalogue
  How much forms is set by the limiting reagent. Once the magnesium is used up, more chlorine cannot make more salt: it has nothing to react with and stays in excess. To make more product, add more of the reactant that ran out.
- Wrong $8$ ⁠ units of a salt richer in chlorine, $\mathrm{MgCl_4}$ ⁠
  Misconception: Thinks the reactant in excess is taken into the product, making it richer in that element.
  Feedback from the catalogue
  The product is always built in the same ratio: every unit of $\mathrm{MgCl_2}$ ⁠ is one magnesium and two chlorine, however much chlorine is in the tank. Whatever finds no partner stays as it is, in excess, and is not part of the product. That is the law of definite proportions.
Once the prediction is locked in, the Scene changes
Run the reaction
Reaction overfixed
The Scene then shows: 8 · Product units made once the reaction has run/nodes/tank/outputs/formed
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
$8$ ⁠ units of $\mathrm{MgCl_2}$ ⁠, as with the exact amount, and $8$ ⁠ chlorine molecules that found no partner. Magnesium limits: each magnesium atom takes exactly two chlorine, and once the magnesium is used up the extra chlorine stays chlorine. The readings also show the mass ratio has not moved: $2.92$ ⁠ grams of chlorine per gram of magnesium.

Worked solution
1. Eight Mg atoms need $8$ ⁠ $\mathrm{Cl_2}$ ⁠ molecules (ratio $1 : 1$ ⁠). There are $16$ ⁠.
2. Magnesium limits: $8$ ⁠ units of $\mathrm{MgCl_2}$ ⁠ form.
3. $16 - 8 = 8$ ⁠ $\mathrm{Cl_2}$ ⁠ molecules are left in excess.
More of a reactant already in excess adds no product and does not change its formula. The product is always $\mathrm{MgCl_2}$ ⁠, because the charges $\mathrm{Mg^{2+}}$ ⁠ and $\mathrm{Cl^-}$ ⁠ fix it.
Step 4 of 6
Open this step
Setting the amount of product
Practice · Reach the target
Question
The tank holds $8$ ⁠ hydrogen molecules, and right now twice the chlorine needed, so $16$ ⁠ $\mathrm{HCl}$ ⁠ molecules form. Change Oxygen or chlorine supplied so that exactly $12$ ⁠ $\mathrm{HCl}$ ⁠ molecules form, then press Check.

Goal
- 12 ± 0.1
  At the start of the step: 16
Setup, choices and feedback
The Scene at this step
Compound
Hydrogen chloridefixed
Oxygen or chlorine supplied
2 × needed
Run the reaction
Reaction overfixed
Read from the Scene at
Product units made once the reaction has run/nodes/tank/outputs/formed
One setting that reaches it
Oxygen or chlorine supplied
0.75 × needed
Feedback for a right answer
At 0.75 of what is needed there are $6$ ⁠ chlorine molecules, and they give $6 \times 2 = 12$ ⁠ $\mathrm{HCl}$ ⁠ molecules. The chlorine runs out first, and $2$ ⁠ hydrogen molecules are left in excess. That is how you control the amount of product: through the limiting reagent.

Feedback for a wrong answer
Not yet. Each $\mathrm{Cl_2}$ ⁠ molecule gives two $\mathrm{HCl}$ ⁠ molecules, as long as there is hydrogen.

Hint
Here the chlorine has to be the limiting reagent. How many $\mathrm{Cl_2}$ ⁠ molecules give $12$ ⁠ $\mathrm{HCl}$ ⁠ molecules?

Worked solution
1. $\mathrm{H_2 + Cl_2 \rightarrow 2HCl}$ ⁠: each $\mathrm{Cl_2}$ ⁠ gives $2$ ⁠ HCl.
2. Twelve HCl need $12 \div 2 = 6$ ⁠ $\mathrm{Cl_2}$ ⁠ molecules.
3. Eight hydrogen molecules need $8$ ⁠, so $6$ ⁠ is $\dfrac{6}{8} = 0.75$ ⁠ of that. Set Oxygen or chlorine supplied to 0.75.
While chlorine limits, each chlorine molecule you add gives two more HCl molecules.
Step 5 of 6
Open this step
The theoretical yield
Practice · Calculate
Question
You mix 8.0 g of hydrogen with 32 g of oxygen and light it. What is the greatest mass of water that can form? ( $M(\mathrm{H_2}) = 2.016$ ⁠, $M(\mathrm{O_2}) = 32.00$ ⁠, $M(\mathrm{H_2O}) = 18.015$ ⁠, in g/mol.)

Correct answer
36.03 g ± 1%
Setup, choices and feedback
The Scene at this step
Compound
Water
Oxygen or chlorine supplied
1 × needed
Run the reaction
Reaction over
Accepted within
± 1%
A fixed value in the question
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 71.5 g ± 1%
  Misconception: Thinks the reactant there is less of (fewer particles, or a smaller mass) is always the one that runs out.
  Feedback from the catalogue
  The limiting reagent is the one that runs out by the ratio in the equation, not the one there is less of. Twelve oxygen molecules need $24$ ⁠ hydrogen molecules, so $16$ ⁠ hydrogen molecules run out first, though there are more of them. Compare what you have with what you need: divide each amount by its coefficient, and the smaller result limits.
- 40 g ± 1%
  Misconception: Thinks the reactant in excess is taken into the product, making it richer in that element.
  Feedback from the catalogue
  The product is always built in the same ratio: every unit of $\mathrm{MgCl_2}$ ⁠ is one magnesium and two chlorine, however much chlorine is in the tank. Whatever finds no partner stays as it is, in excess, and is not part of the product. That is the law of definite proportions.
Feedback for a right answer
36.0 g. There are $3.97$ ⁠ mol of hydrogen and $1.00$ ⁠ mol of oxygen. The oxygen needs only $2.00$ ⁠ mol of hydrogen, so oxygen limits, though there is four times as much of it in grams. Two moles of water weigh $36.0$ ⁠ grams, and 4.0 g of hydrogen is left.

Hint
Turn both reactants into moles, and see which runs out first by $\mathrm{2H_2 + O_2 \rightarrow 2H_2O}$ ⁠.

Worked solution
1. Moles: $n(\mathrm{H_2}) = \dfrac{8.0}{2.016} = 3.97$ ⁠; $n(\mathrm{O_2}) = \dfrac{32}{32.00} = 1.00$ ⁠.
2. By the equation $\mathrm{H_2 : O_2 = 2 : 1}$ ⁠. One mole of $\mathrm{O_2}$ ⁠ needs $2.00$ ⁠ mol of $\mathrm{H_2}$ ⁠, and there are $3.97$ ⁠: oxygen limits.
3. $n(\mathrm{H_2O}) = 2 \times n(\mathrm{O_2}) = 2.00$ ⁠.
4. $m(\mathrm{H_2O}) = 2.00 \times 18.015 = 36.0$ ⁠ grams.
If you got $71.5$ ⁠, you chose hydrogen as limiting because there are fewer grams of it. If you got $40$ ⁠, you added up all the reactants, including the hydrogen left in excess.
Step 6 of 6
Open this step
Percentage yield
Practice · Calculate
Question
A 12 g magnesium ribbon should give 47.0 g of $\mathrm{MgCl_2}$ ⁠. In the lab you collected 40.0 g. What is the percentage yield? (Answer in per cent.)

Correct answer
85.1 ± 0.5
Setup, choices and feedback
The Scene at this step
Compound
Magnesium chloride
Oxygen or chlorine supplied
1 × needed
Run the reaction
Reaction over
Accepted within
± 0.5
A fixed value in the question
Wrong answers it recognises
- 117.5 ± 0.5
  Misconception: Divides the theoretical yield by the actual yield, giving a percentage above 100.
  Feedback from the catalogue
  The percentage yield is what you actually got over what you should have got, times $100$ ⁠: $\dfrac{m_{\text{actual}}}{m_{\text{theoretical}}} \times 100$ ⁠. You cannot get more product than the atoms allow, so a result over $100\%$ ⁠ says the fraction is upside down.
Feedback for a right answer
$\dfrac{40.0}{47.0} \times 100 = 85.1\%$ ⁠. No atoms were lost: some salt stayed on the glassware or went astray in transfer, and some magnesium may not have reacted completely.

Hint
What you got over what you should have got.

Worked solution
1. The theoretical yield (from the calculation): $47.0$ ⁠ grams.
2. The actual yield (from the balance): $40.0$ ⁠ grams.
3. The percentage yield: $\dfrac{40.0}{47.0} \times 100 = 85.1\%$ ⁠.
If you got $117.5\%$ ⁠, you divided the wrong way round. A yield cannot pass $100\%$ ⁠: you cannot get more salt than the magnesium allows.

Across the unit · 16 misconceptions

Misconception catalogue

Each common mistake this unit recognises, in a few words. Open one for what the learner reads when an answer matches it, and where it is asked.

Index

Conservation of mass

Atoms change partners

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

Counting atoms

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Weighing the tank

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

Open or sealed

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

A metal that gains weight

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Balancing equations

Coefficients

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

Two from one

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Just enough chlorine

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

Formulas set by charges

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

Balancing by counting

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Reacting masses

Relative molecular mass

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution