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GCSE Chemistry · GCSEChapter 8 · The mole, reacting masses and limiting reactants
GCSE · Chapter 8

The mole, reacting masses and limiting reactants

By the end of this chapter you will be able to

  • Calculate the number of moles from a mass and a Mr, and the mass from the number of moles
  • Use moles and a balanced equation to find the mass of a product
  • Calculate a percentage yield, and an atom economy
  • Find the limiting reactant from amounts of two reactants
  • [H] Calculate the volume of a gas from the number of moles

You should already know

  • Relative formula mass (Chapter 7)
  • Balancing equations and mass ratios (Chapter 7)

Where this leads

  • Concentration of solutions in mol/dm3 (Chapter 9)
  • Acids, alkalis and titrations (Chapter 10)
  • Electrolysis and Faraday (Chapter 11)

GCSE Chemistry. Content follows the GCSE (9-1) subject content for England (Edexcel and AQA). Content references: Use the mole and the Avogadro constant, and calculate moles from mass and mass from moles, Calculate the masses of reactants and products from a balanced equation, Calculate percentage yield and atom economy, Identify the limiting reactant and the excess reactant, [H] Calculate the volume of a gas at room temperature and pressure from moles (24 dm3 per mole).

Exercise P  Do you remember?

1
FWhat is 36 / 18? Write the number.
2
FWhat is 0.5 x 44? Write the number.
3
FWhat is 2 x 6 + 2 x 1? Write the number.
4
FWrite 2.5 x 10-1 as a decimal. Write the number.
5
FWhat is 4.8 / 24? Write the number.
8.1

The mole and the Avogadro constant

GCSE
Use the mole and the Avogadro constant, and calculate moles from mass and mass from moles
Video lesson: The mole and the Avogadro constantComing soon
Context. A technician needs to weigh out a precise amount of salt for an experiment. “One mole of salt weighs its Mr in grams,” she says. A mole is a counting unit, like a dozen, and it lets chemists count particles by weighing them.

A mole is the amount of substance that contains the Avogadro constant, 6.02 x 1023, of particles. The mass of one mole of a substance in grams equals its Mr (numerically). So the number of moles is:

moles = mass (g) / Mr    and    mass (g) = moles x Mr

QuantityFormulaExample
molesmass (g) / Mr36 g of water: 36 / 18 = 2 mol
mass (g)moles x Mr0.5 mol of CO2: 0.5 x 44 = 22 g
particlesmoles x 6.02 x 10232 mol of water: 2 x 6.02 x 1023 = 1.204 x 1024
The mole calculations

Key ideas

  • One mole is 6.02 x 1023 particles (atoms, molecules or ions).
  • The mass of one mole, in grams, equals the Mr.
  • moles = mass / Mr; mass = moles x Mr.
  • Write the formula first, so you can find the Mr.
Worked example 8.1.1
Calculate the number of moles in 11.7 g of sodium chloride, NaCl (Mr 58.5). Write the number.

Method. Divide the mass by the Mr.

  1. Mr. The Mr of NaCl is 58.5.
    58.5
  2. Divide. moles = 11.7 / 58.5.
    11.7 / 58.5
  3. Answer. So the number of moles is 0.2.
    0.2
  4. Write the answer.
    0.2

Check. 0.2 x 58.5 = 11.7. ✓

Worked example 8.1.2
Calculate the mass of 0.5 moles of carbon dioxide, CO2 (Mr 44). Write the number in g.

Method. Multiply the moles by the Mr.

  1. Mr. The Mr of CO2 is 44.
    44
  2. Multiply. mass = 0.5 x 44.
    0.5 x 44
  3. Answer. So the mass is 22 g.
    22
  4. Write the answer.
    22

Check. 22 / 44 = 0.5 mol. ✓

Worked example 8.1.3
How many molecules are in 2 moles of water? Write the number in the form a x 10b.

Method. Multiply the moles by the Avogadro constant.

  1. Constant. One mole contains 6.02 x 1023 molecules.
    6.02 x 1023
  2. Multiply. 2 x 6.02 x 1023 = 12.04 x 1023.
    12.04 x 1023
  3. Standard form. In standard form this is 1.204 x 1024.
    1.204 x 1024
  4. Write the answer.
    1.204 x 10^24

Check. Check: 1.204 x 1024 / 6.02 x 1023 = 2. ✓

Watch out

Watch out: the mole is a number of particles, not a mass. Always divide by the Mr (not the Ar of one atom) when the substance is a compound.

Exam tip

Exam tip. Write the three numbers in every mole question: mass, Mr and moles. Then check the units: grams in, moles out.

Exercise 8A  The mole and the Avogadro constant

1
FWhat is the Avogadro constant? Write the number in the form a x 10b.
2
FHow many moles are in 36 g of water, H2O (Mr 18)? Write the number.
3
FHow many moles are in 40 g of MgO (Mr 40)? Write the number.
4
FWhat is the mass of 1 mole of CaCO3 (Mr 100)? Write the number in g.
5
FWhat is the mass of 3 moles of MgO (Mr 40)? Write the number in g.
6
FWhat is the mass of 0.5 moles of CO2 (Mr 44)? Write the number in g.
7
FHow many moles are in 11.7 g of NaCl (Mr 58.5)? Write the number.
8
FHow many moles are in 4.8 g of magnesium, Mg (Ar 24)? Write the number.
9
FWhat is the mass of 0.25 moles of MgCl2 (Mr 95)? Write the number in g.
10
FHow many molecules are in 1 mole of H2O? Write the number in the form a x 10b.
11
PSHow many moles are in 1.8 g of water (Mr 18)? Write the number.
12
PSHow many molecules are in 0.5 mol of CO2? Write the number in the form a x 10b.
13
RExplain why the mass of one mole of a compound depends on the compound, in one sentence. Accept any sentence that says different compounds have different Mr values.
14
RA pupil says “one mole of water has the same mass as one mole of carbon dioxide.” Is this right? Write yes or no, and give one reason.
8.2

Reacting masses from balanced equations

GCSE
Calculate the masses of reactants and products from a balanced equation
Video lesson: Reacting masses from balanced equationsComing soon
Context. A school lab wants to know how much magnesium chloride can be made from a strip of magnesium. “Use the equation,” says the teacher, “and turn the grams into moles first.” This is the method for every reacting mass question.

The method has four steps: write the balanced equation; work out the moles of the substance you know (mass / Mr); use the ratio from the equation to find the moles of the substance you want; then change moles to mass (moles x Mr).

mass of A (g)moles of Amoles of B/ Mrratiothen multiply by the Mr of B to get the mass of B
Reacting masses: from the mass you know to the mass you want

Key ideas

  • Balanced equation first: the ratio of moles comes from the numbers in front.
  • Moles = mass / Mr; mass = moles x Mr.
  • Ratio: for 2Mg + O2 -> 2MgO, 2 mol Mg gives 2 mol MgO.
  • Keep the number of decimal places sensible; give the final answer to 2 or 3 significant figures.
Worked example 8.2.1
Magnesium reacts with hydrochloric acid: Mg + 2HCl -> MgCl2 + H2. What mass of MgCl2 is made from 6.0 g of Mg? Write the number in g.

Method. Use the four steps.

  1. Moles of Mg. moles = 6.0 / 24 = 0.25 mol.
    0.25 mol
  2. Ratio. The equation has 1 Mg for 1 MgCl2, so 0.25 mol MgCl2.
    0.25 mol
  3. Mass. mass = 0.25 x 95 = 23.75 g.
    23.75 g
  4. Write the answer.
    23.75

Check. Check: 23.75 / 95 = 0.25 mol. ✓

Worked example 8.2.2
Calcium carbonate decomposes: CaCO3 -> CaO + CO2. What mass of CO2 is made from 50 g of CaCO3? Write the number in g.

Method. Find the moles of CaCO3, then the moles of CO2.

  1. Moles. moles of CaCO3 = 50 / 100 = 0.5 mol.
    0.5 mol
  2. Ratio. 1 CaCO3 gives 1 CO2, so 0.5 mol CO2.
    0.5 mol
  3. Mass. mass = 0.5 x 44 = 22 g.
    22 g
  4. Write the answer.
    22

Check. The mass of CaO is 28 g, and 22 + 28 = 50 g. ✓

Worked example 8.2.3
In 2Mg + O2 -> 2MgO, what mass of oxygen reacts with 12 g of magnesium? Write the number in g.

Method. Find the moles of Mg, then the moles of O2.

  1. Moles of Mg. moles = 12 / 24 = 0.5 mol.
    0.5 mol
  2. Ratio. 2 Mg react with 1 O2, so 0.5 mol Mg needs 0.25 mol O2.
    0.25 mol
  3. Mass. mass of O2 = 0.25 x 32 = 8 g.
    8 g
  4. Write the answer.
    8

Check. 12 g Mg + 8 g O2 = 20 g MgO. ✓

Watch out

Watch out: use the numbers in front of the formulae for the ratio, not the subscripts. In Mg + 2HCl, 1 Mg reacts with 2 HCl, so for every mole of Mg you need two moles of HCl.

Exam tip

Exam tip. Write the balanced equation at the top of your working, and label each line with the substance you are finding.

Exercise 8B  Reacting masses from balanced equations

1
FIn 2H2 + O2 -> 2H2O, how many moles of H2O are made from 2 moles of H2? Write the number.
2
FIn Mg + 2HCl -> MgCl2 + H2, how many moles of HCl react with 1 mole of Mg? Write the number.
3
FIn CaCO3 -> CaO + CO2, how many moles of CO2 are made from 1 mole of CaCO3? Write the number.
4
FIn 2Mg + O2 -> 2MgO, how many moles of MgO are made from 2 moles of Mg? Write the number.
5
FWhat is the mass of 0.5 mol of Mg (Ar 24)? Write the number in g.
6
FHow many moles of Mg are in 12 g of magnesium? Write the number.
7
PSIn Mg + 2HCl -> MgCl2 + H2, what mass of MgCl2 is made from 0.25 mol of Mg? Write the number in g.
8
PSIn 2Mg + O2 -> 2MgO, what mass of MgO is made from 4.8 g of Mg? Write the number in g.
9
PSIn 2H2 + O2 -> 2H2O, what mass of H2O is made from 0.5 mol of O2? Write the number in g.
10
PSWhat mass of oxygen, O2, is needed to react with 0.5 mol of Mg in 2Mg + O2 -> 2MgO? Write the number in g.
11
RA pupil says the equation Mg + HCl -> MgCl2 + H2 can be used as it is for calculations. Explain why this is wrong, in one sentence. Accept any sentence that says the equation must be balanced first, so that the ratio of moles is correct.
12
RExplain why the mass of product can be calculated from the mass of reactant using moles, in one sentence. Accept any sentence that says the ratio in the balanced equation links the moles of each substance.
8.3

Percentage yield and atom economy

GCSE
Calculate percentage yield and atom economy
Video lesson: Percentage yield and atom economyComing soon
Context. A student makes copper oxide from copper, but the mass of copper oxide she collects is less than the teacher expected. “Why did I lose some product?” she asks. The answer is the difference between the theoretical yield and the actual yield.

The theoretical yield is the maximum mass of product calculated from the equation. The actual yield is the mass actually obtained. The percentage yield = actual / theoretical x 100. In a chemical plant, a high yield saves money; in the lab, losses come from transfers, side reactions and reactions that do not go to completion.

The atom economy measures how much of the reactant atoms end up in the wanted product: atom economy = Mr of the wanted product / total Mr of all reactants x 100.

QuantityFormulaExample
percentage yieldactual / theoretical x 1006.4 g from 8.0 g expected: 80%
atom economyMr of wanted product / total Mr of reactants x 100CaO from CaCO3: 56 / 100 x 100 = 56%
Percentage yield and atom economy

Key ideas

  • Percentage yield = actual / theoretical x 100.
  • A yield can never be more than 100% in a sensible experiment; losses reduce the yield.
  • Atom economy = Mr of the wanted product / total Mr of reactants x 100.
  • Atom economy is 100% when every reactant atom ends up in the wanted product.
Worked example 8.3.1
A reaction should make 8.0 g of product, but 6.4 g is obtained. What is the percentage yield? Write the number.

Method. Divide the actual by the theoretical yield.

  1. Actual. Actual yield = 6.4 g.
    6.4 g
  2. Theoretical. Theoretical yield = 8.0 g.
    8.0 g
  3. Percentage. 6.4 / 8.0 x 100 = 80.
    80
  4. Write the answer.
    80

Check. 80% of 8.0 g is 6.4 g. ✓

Worked example 8.3.2
In CaCO3 -> CaO + CO2, what is the atom economy for making CaO? Write the number as a percentage.

Method. Use the Mr of CaO and the total Mr of the reactants.

  1. Wanted product. Mr of CaO = 56.
    56
  2. Reactant. Mr of CaCO3 = 100.
    100
  3. Atom economy. 56 / 100 x 100 = 56.
    56
  4. Write the answer.
    56

Check. The CO2 is the waste product, so the economy is less than 100%. ✓

Worked example 8.3.3
In 2H2 + O2 -> 2H2O, what is the atom economy for making water? Write the number as a percentage.

Method. Compare the Mr of the wanted product with all the reactants.

  1. Wanted product. 2H2O has a total Mr of 36.
    36
  2. Reactants. 2H2 (4) + O2 (32) = 36.
    36
  3. Atom economy. 36 / 36 x 100 = 100.
    100
  4. Write the answer.
    100

Check. All the atoms end up in water, so there is no waste. ✓

Watch out

Watch out: percentage yield uses the actual and theoretical yields; atom economy uses the Mr values of the equation. They are different ideas.

Exam tip

Exam tip. A low percentage yield does not mean the reaction is wrong; it may be lost in transfers. Give the reason, such as a spill or an incomplete reaction.

Exercise 8C  Percentage yield and atom economy

1
FWhat is the theoretical yield? Write the words.
2
FWhat is the percentage yield when 9.0 g is obtained from a theoretical 10.0 g? Write the number.
3
FWhat is the percentage yield when 6.4 g is obtained from a theoretical 8.0 g? Write the number.
4
FWhat is the atom economy of 2H2 + O2 -> 2H2O for making water? Write the number as a percentage.
5
FWhat is the atom economy of CaCO3 -> CaO + CO2 for making CaO? Write the number as a percentage.
6
PSA reaction should make 20 g, but 15 g is collected. What is the percentage yield? Write the number.
7
PSIn a reaction with a theoretical yield of 50 g, the percentage yield is 60%. What mass of product was obtained? Write the number in g.
8
PSIn a reaction, the theoretical yield is 12.0 g and the actual yield is 9.6 g. Find the percentage yield. Write the number.
9
PSCalculate the atom economy of making MgO from 2Mg + O2 -> 2MgO. Write the number as a percentage.
10
RExplain why a reaction can have a 100% atom economy but a low percentage yield, in one sentence. Accept any sentence that says atom economy uses the equation, while percentage yield depends on losses in the experiment.
11
RExplain why companies want a high atom economy, in one sentence. Accept any sentence that says less waste is made, which saves raw materials and money.
8.4

Limiting reactants

GCSE
Identify the limiting reactant and the excess reactant
Video lesson: Limiting reactantsComing soon
Context. A pupil mixes 0.3 mol of magnesium with 0.1 mol of oxygen and heats the mixture. “Which one will run out first?” she asks. The limiting reactant is the one that is used up first; the other is in excess.

In a reaction with two reactants, the limiting reactant is the one that is used up first, so it limits the amount of product. The other reactant is in excess and some of it is left over. To find the limiting reactant, use the equation to find how much of the other reactant is needed for all of one reactant.

ReactantMoles availableMoles needed for all of the otherResult
magnesium0.30.2 (to react with all 0.1 mol O2)excess: 0.1 mol left over
oxygen0.10.15 (to react with all 0.3 mol Mg)limiting: it runs out first
Finding the limiting reactant in 2Mg + O2 -> 2MgO, with 0.3 mol Mg and 0.1 mol O2

Key ideas

  • Limiting reactant: used up first; it limits the mass of product.
  • Excess reactant: some is left over at the end of the reaction.
  • Method: work out the moles of each reactant, then use the ratio to see which runs out.
  • The mass of product is worked out from the limiting reactant.
Worked example 8.4.1
0.3 mol Mg reacts with 0.1 mol O2 in 2Mg + O2 -> 2MgO. Which is the limiting reactant? Write the formula.

Method. Find the O2 needed for all the Mg.

  1. Needed. 0.3 mol Mg needs 0.3 / 2 = 0.15 mol O2.
    0.15 mol
  2. Available. Only 0.1 mol O2 is available, which is less than 0.15.
    0.1 < 0.15
  3. Decide. So O2 runs out first and is the limiting reactant.
    O2
  4. Write the answer.
    O2

Check. Magnesium is in excess. ✓

Worked example 8.4.2
In the same reaction, what mass of MgO is made? Write the number in g.

Method. Use the limiting reactant O2.

  1. Moles. 0.1 mol O2 makes 2 x 0.1 = 0.2 mol MgO.
    0.2 mol
  2. Mass. mass = 0.2 x 40 = 8.0 g.
    8.0
  3. Write the answer.
    8

Check. Use the limiting reactant to find the product. ✓

Worked example 8.4.3
In the same reaction, what mass of magnesium is left over? Write the number in g.

Method. Find the Mg used, then subtract from the Mg present.

  1. Used. The 0.1 mol O2 reacts with 2 x 0.1 = 0.2 mol Mg.
    0.2 mol
  2. Left over. Excess = 0.3 - 0.2 = 0.1 mol.
    0.1 mol
  3. Mass. mass = 0.1 x 24 = 2.4 g.
    2.4
  4. Write the answer.
    2.4

Check. The leftover magnesium is the excess reactant. ✓

Watch out

Watch out: do not decide the limiting reactant by the larger number of moles. Compare each reactant with what the equation says it needs.

Exam tip

Exam tip. Write the ratio from the equation, then check whether there is enough of each reactant for the other. Say which one runs out.

Exercise 8D  Limiting reactants

1
FWhat is the limiting reactant? Write the words.
2
FWhat is an excess reactant? Write the words.
3
FIn 2Mg + O2 -> 2MgO, how many moles of O2 are needed for 0.2 mol Mg? Write the number.
4
FIn 2Mg + O2 -> 2MgO, how many moles of O2 are needed for 0.3 mol Mg? Write the number.
5
PSIn 2Mg + O2 -> 2MgO, with 0.3 mol Mg and 0.2 mol O2, which is the limiting reactant? Write the formula.
6
PSIn 2Mg + O2 -> 2MgO, with 0.3 mol Mg and 0.1 mol O2, how many moles of MgO are made? Write the number.
7
PSIn Mg + 2HCl -> MgCl2 + H2, 0.2 mol Mg reacts with 0.3 mol HCl. Which is limiting? Write the formula.
8
PSIn Mg + 2HCl -> MgCl2 + H2, with 0.2 mol Mg and 0.3 mol HCl, how many moles of MgCl2 are made? Write the number.
9
PSIn 2Mg + O2 -> 2MgO, 0.3 mol Mg reacts with 0.1 mol O2. What mass of magnesium is left over? Write the number in g.
10
RExplain why the mass of product depends on the limiting reactant, in one sentence. Accept any sentence that says the product stops forming when the limiting reactant runs out.
11
RExplain why using more of one reactant does not always give more product, in one sentence. Accept any sentence that says the other reactant is limiting and cannot make more product.
8.5

Volumes of gases [H]

GCSE
[H] Calculate the volume of a gas at room temperature and pressure from moles (24 dm3 per mole)
Video lesson: Volumes of gases [H]Coming soon
Context. A pupil collects hydrogen gas in a gas syringe from a reaction with magnesium. “How much gas should I expect?” she asks. If the amount of gas in moles is known, its volume can be calculated at room temperature and pressure.

[H] Higher tier only. At room temperature and pressure (about 20 degrees C and 1 atmosphere), one mole of any gas occupies 24 dm3. So volume (dm3) = moles x 24. For example, 0.5 mol of CO2 occupies 12 dm3.

Moles of gasVolume at room temperature and pressure (dm3)
0.512
0.24.8
124
248
Volume of gases at room temperature and pressure

Key ideas

  • Volume of gas (dm3) = moles x 24 at room temperature and pressure.
  • Moles = volume (dm3) / 24.
  • The same number of moles of any gas gives the same volume.
  • Use the balanced equation to find the moles of gas before you calculate its volume.
Worked example 8.5.1
[H] What volume does 0.2 mol of hydrogen occupy at room temperature and pressure? Write the number in dm3.

Method. Multiply the moles by 24.

  1. Multiply. volume = 0.2 x 24 = 4.8.
    4.8
  2. Units. The volume is in dm3.
    dm3
  3. Write the answer.
    4.8

Check. 4.8 / 24 = 0.2 mol. ✓

Worked example 8.5.2
[H] In Mg + 2HCl -> MgCl2 + H2, 0.2 mol of Mg reacts. What volume of hydrogen is made at room temperature and pressure? Write the number in dm3.

Method. Use the ratio Mg : H2 = 1 : 1.

  1. Ratio. 0.2 mol Mg gives 0.2 mol H2.
    0.2 mol
  2. Volume. volume = 0.2 x 24 = 4.8 dm3.
    4.8
  3. Write the answer.
    4.8

Check. The same as the gas in the first example. ✓

Worked example 8.5.3
[H] What volume is occupied by 2 mol of a gas? Write the number in dm3.

Method. Multiply by 24.

  1. Multiply. 2 x 24 = 48.
    48
  2. Units. The answer is in dm3.
    dm3
  3. Write the answer.
    48

Check. One mole is 24 dm3, so two moles is 48 dm3. ✓

Watch out

Watch out: 24 dm3 is only at room temperature and pressure. It is not the volume of a mole of gas at other conditions.

Exam tip

Exam tip. Write the moles first, then multiply by 24. Give the unit dm3.

Exercise 8E  Volumes of gases [H]

1
F[H] What is the volume of one mole of any gas at room temperature and pressure? Write the number in dm3.
2
F[H] What is the volume of 1 mol of oxygen gas? Write the number in dm3.
3
F[H] What is the volume of 0.5 mol of CO2? Write the number in dm3.
4
F[H] How many moles are in 48 dm3 of gas at room temperature and pressure? Write the number.
5
F[H] How many moles are in 4.8 dm3 of gas at room temperature and pressure? Write the number.
6
PS[H] What volume of hydrogen is made from 0.25 mol of Mg in Mg + 2HCl -> MgCl2 + H2? Write the number in dm3.
7
PS[H] What volume of CO2 is made from 50 g of CaCO3? (Mr 100, and 1 mol CO2 = 24 dm3.) Write the number in dm3.
8
PS[H] A gas syringe collects 4.8 dm3 of hydrogen. How many moles of hydrogen are collected? Write the number.
9
R[H] Explain why a volume of 24 dm3 per mole is used for gases but not for solids, in one sentence. Accept any sentence that says the volume of a solid depends on its density, while gas particles are far apart and the molar volume is the same for all gases.

Chapter summary

SkillMethod
Mole6.02 x 1023 particles; the mass of one mole in grams equals the Mr.
Mass and molesmoles = mass / Mr; mass = moles x Mr.
Reacting massesBalanced equation -> moles of known -> ratio -> moles of wanted -> mass.
Percentage yieldactual / theoretical x 100.
Atom economyMr of wanted product / total Mr of reactants x 100.
Limiting reactantThe reactant used up first; it sets the amount of product.
[H] Gas volumeVolume (dm3) = moles x 24 at room temperature and pressure.

Can you…?

  • I can calculate moles from mass and Mr, and mass from moles
  • I can use a balanced equation to calculate the mass of a product
  • I can calculate percentage yield and atom economy
  • I can identify the limiting reactant and find the excess left over
  • [H] I can calculate the volume of a gas from its number of moles

Key terms

mole
the amount of substance that contains 6.02 x 1023 particles
Avogadro constant
6.02 x 1023 particles per mole
theoretical yield
the maximum mass of product calculated from the equation
percentage yield
the actual yield as a percentage of the theoretical yield
atom economy
the percentage of the reactant mass that ends up in the wanted product
limiting reactant
the reactant that is used up first and limits the product
excess reactant
the reactant that is left over at the end of a reaction

Exercise 8Z  Check-up

1
How many moles are in 58.5 g of NaCl (Mr 58.5)? Write the number.
2
What is the mass of 2 mol of water (Mr 18)? Write the number in g.
3
What is the Avogadro constant? Write the number in the form a x 10b.
4
What is the mass of MgO produced from 0.2 mol of MgO (Mr 40)? Write the number in g.
5
In CaCO3 -> CaO + CO2, how many moles of CO2 are made from 1 mol CaCO3? Write the number.
6
A reaction gives 16 g of product, but the theoretical yield is 20 g. What is the percentage yield? Write the number.
7
What is the atom economy of CaCO3 -> CaO + CO2 for CaO? Write the number as a percentage.
8
In 2Mg + O2 -> 2MgO, with 0.3 mol Mg and 0.1 mol O2, which is the limiting reactant? Write the formula.
9
[H] What volume does 0.5 mol of hydrogen occupy at room temperature and pressure? Write the number in dm3.
10
Explain why the mass of product is calculated from the limiting reactant, in one sentence. Accept any sentence that says the product stops forming when the limiting reactant is used up.

Total 25 marks. Allow about 25 minutes.

Exercise 8X  Exam-style questions

1
FCalculate the number of moles in 4.8 g of magnesium (Ar 24). Write the number.[1 mark]
2
PSIn 2Mg + O2 -> 2MgO, what mass of MgO is made from 4.8 g of Mg? Write the number in g.[3 marks]
3
PSCalculate the percentage yield when 6.4 g of product is made from a theoretical yield of 8.0 g. Write the number.[2 marks]
4
PSCalculate the atom economy for making CaO from CaCO3. Write the number as a percentage.[2 marks]
5
PSIn Mg + 2HCl -> MgCl2 + H2, a pupil uses 0.2 mol Mg and 0.3 mol HCl. Which reactant is limiting? Write the formula, with a reason.[3 marks]
6
PSIn 2Mg + O2 -> 2MgO, what mass of oxygen is needed to react with 12 g of magnesium? Write the number in g.[3 marks]
7
RExplain why the actual yield of a reaction is usually lower than the theoretical yield, in one or two sentences. Accept any answer that gives one loss, such as losses in transfers, or a reaction that does not go to completion.[3 marks]
8
PS[H] Calculate the volume of hydrogen made from 0.25 mol of Mg in Mg + 2HCl -> MgCl2 + H2, at room temperature and pressure. Write the number in dm3.[3 marks]
9
PSA pupil heats 50 g of CaCO3. What mass of CaO is made, if the reaction is complete? Write the number in g.[3 marks]
10
PSA calculation predicts 15 g of product from a reaction, but 12 g is collected. Calculate the percentage yield. Write the number.[2 marks]
← Chapter 7: Relative formula mass and conservation of massChapter 9: Concentration of solutions →

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