Answers · Concentration, gas volumes and yield

How much gas will you get? — answer key

Check meaning, not exact wording. These are teaching points, not official Cambridge marks.

1. YQ1 · Practice

Model answer. At r.t.p. one mole of any gas takes up 24 dm³, which is 24 000 cm³: the molar gas volume. Gas particles are far apart, so the number of particles, not their size or mass, decides the volume.

A full answer includes

24 dm³ (24 000 cm³), whatever the gas.

Also accept

24 dm³

Watch for

“One mole of gas is 24 cm³.” — It is 24 dm³, which is 24 000 cm³. For a volume in cm³, use 24 000.
“A heavier gas takes up more space.” — A mole of any gas takes up 24 dm³ at r.t.p. The particles are far apart, so their size and mass hardly matter.

2. YQ2 · Practice

Model answer. (a) Each mole takes up 24 dm³, so 0.250 mol takes up 0.250 × 24 = 6.00 dm³. (b) The volume is in cm³, so divide by 24 000 cm³: 120 ÷ 24 000 = 0.00500 mol.

Checked automatically

(a) Volume of oxygen: 6.00 dm³ (3 significant figures)
- Common wrong answer 0.0104 : You divided. Each mole takes up 24 dm³, so multiply: 0.250 × 24.
(b) Amount of hydrogen: 0.00500 mol (3 significant figures)
- Common wrong answer 5.00 : You divided 120 cm³ by 24. A mole is 24 000 cm³, so divide by 24 000.

A full answer includes

(a) V = 0.250 × 24 = 6.00 dm³.
(b) n = 120 ÷ 24 000 = 0.00500 mol.

Also accept

6.00 dm³; 0.00500 mol

Watch for

“One mole of gas is 24 cm³.” — It is 24 dm³, which is 24 000 cm³. For a volume in cm³, use 24 000.

3. YQ3 · Practice

Model answer. The equation reads in moles: 2 mol of H₂ react with 1 mol of O₂ and make 2 mol of H₂O. Oxygen is half the hydrogen, 0.200 mol, and water is the same as the hydrogen, 0.400 mol.

Checked automatically

(a) Oxygen needed: 0.200 mol (3 significant figures)
- Common wrong answer 0.800 : The ratio H₂ : O₂ is 2 : 1, so the oxygen is half the hydrogen, not double.
(b) Water formed: 0.400 mol (3 significant figures)
- Common wrong answer 0.800 : H₂ : H₂O is 2 : 2, which is 1 : 1: the same amount as the hydrogen.
- Common wrong answer 0.200 : H₂ : H₂O is 2 : 2, which is 1 : 1: the same amount as the hydrogen.

A full answer includes

(a) 0.400 ÷ 2 = 0.200 mol of O₂.
(b) 0.400 mol of H₂O (the ratio H₂ : H₂O is 2 : 2).

Also accept

0.200 mol; 0.400 mol

Watch for

“Multiply by the number in front of the substance you want.” — Use the ratio of the two numbers: O₂ : H₂ is 1 : 2, so halve.

4. YQ4 · Practice

Model answer. M(CuO) = 64 + 16 = 80 g/mol, so n = 8.00 ÷ 80 = 0.100 mol. One CuO gives one Cu, so 0.100 mol of copper forms: 0.100 × 64 = 6.40 g. The oxygen has left the copper, so the copper is lighter than the oxide.

Checked automatically

Mass of copper: 6.40 g (3 significant figures)
- Common wrong answer 8.00 : The mass does not stay the same from reactant to product. Go through moles.
- Common wrong answer 0.100 : That is the amount of copper in mol. Multiply by M = 64 g/mol for the mass.

A full answer includes

n(CuO) = 8.00 ÷ 80 = 0.100 mol.
Ratio CuO : Cu = 1 : 1, so 0.100 mol of Cu.
m = 0.100 × 64 = 6.40 g.

Also accept

6.40 g

Watch for

“The product has the same mass as the reactant.” — Only the total mass is conserved. Go through moles: change to moles, use the equation, change back.

5. YQ5 · Practice

Model answer. n(Ca) = 2.00 ÷ 40 = 0.0500 mol. 2Ca make 2CaO, a ratio of 1 : 1, so 0.0500 mol of CaO forms. M(CaO) = 40 + 16 = 56 g/mol, so m = 0.0500 × 56 = 2.80 g.

Checked automatically

Mass of calcium oxide: 2.80 g (3 significant figures)
- Common wrong answer 5.60 : Ca : CaO is 2 : 2, which is 1 : 1. The amount stays 0.0500 mol; you doubled it.
- Common wrong answer 1.40 : Ca : CaO is 2 : 2, which is 1 : 1. The amount stays 0.0500 mol; you halved it.

A full answer includes

n(Ca) = 2.00 ÷ 40 = 0.0500 mol.
Ratio 2 : 2 = 1 : 1, so 0.0500 mol of CaO.
m = 0.0500 × 56 = 2.80 g.

Also accept

2.80 g

Watch for

“The 2 in front of CaO doubles the answer.” — Both sides have a 2: Ca : CaO = 2 : 2 = 1 : 1.

6. YQ6 · Practice

Model answer. n(Zn) = 0.325 ÷ 65 = 0.00500 mol. One Zn gives one H₂, so 0.00500 mol of hydrogen forms: 0.00500 × 24 000 = 120 cm³. The 2 belongs to the HCl; the acid is in excess, so all the zinc reacts.

Checked automatically

Volume of hydrogen: 120 cm³ (3 significant figures)
- Common wrong answer 0.120 : That number is in dm³. For cm³, multiply the moles by 24 000.
- Common wrong answer 240 : Zn : H₂ is 1 : 1. The 2 in the equation belongs to HCl.

A full answer includes

n(Zn) = 0.325 ÷ 65 = 0.00500 mol.
Zn : H₂ = 1 : 1, so 0.00500 mol of H₂.
V = 0.00500 × 24 000 = 120 cm³.

Also accept

120 cm³

Watch for

“One mole of gas is 24 cm³.” — It is 24 dm³, which is 24 000 cm³. For a volume in cm³, use 24 000.

7. YQ7 · Practice

Model answer. Start from what you know, the gas: n(CO₂) = 1.20 dm³ ÷ 24 dm³/mol = 0.0500 mol. The ratio CaCO₃ : CO₂ is 1 : 1, so you need 0.0500 mol of calcium carbonate. M(CaCO₃) = 100 g/mol, so m = 0.0500 × 100 = 5.00 g.

Checked automatically

Mass of calcium carbonate: 5.00 g (3 significant figures)
- Common wrong answer 28.8 : You multiplied the volume by 24. To find moles, divide: 1.20 ÷ 24.
- Common wrong answer 2880 : You multiplied the volume by 24. To find moles, divide: 1.20 ÷ 24.
- Common wrong answer 2.80 : You used M = 56, the molar mass of CaO. The reactant is CaCO₃, M = 100.

A full answer includes

n(CO₂) = 1.20 ÷ 24 = 0.0500 mol.
CaCO₃ : CO₂ = 1 : 1, so 0.0500 mol of CaCO₃.
m = 0.0500 × 100 = 5.00 g.

Also accept

5.00 g

Watch for

“Multiply a gas volume by 24 to get moles.” — Each mole takes up 24 dm³, so the number of moles is the volume divided by 24.

8. YQ8 · Practice

Model answer. n(CaCO₃) = 0.500 ÷ 100 = 0.00500 mol. Each CaCO₃ needs two HCl, so 0.0100 mol of HCl. Rearrange amount = concentration × volume, n = c × V: V = n ÷ c = 0.0100 ÷ 0.200 = 0.0500 dm³, which is 50.0 cm³.

Checked automatically

Volume of acid: 50.0 cm³ (3 significant figures)
- Common wrong answer 25.0 : Each CaCO₃ needs 2 HCl: double the moles before you divide by c.
- Common wrong answer 0.0500 : That number is in dm³. Multiply by 1000 for cm³.

A full answer includes

n(CaCO₃) = 0.500 ÷ 100 = 0.00500 mol.
CaCO₃ : HCl = 1 : 2, so 0.0100 mol of HCl.
V = 0.0100 ÷ 0.200 = 0.0500 dm³ = 50.0 cm³.

Also accept

50.0 cm³

Watch for

“Use the mole ratio on the volumes of solutions.” — A volume of solution depends on its concentration too. Change to moles, use the equation, then V = n ÷ c.

9. YQ9 · Practice

Model answer. Equal volumes of gases hold equal numbers of moles, so the volumes of the gases come in the ratio C₃H₈ : O₂ : CO₂ = 1 : 5 : 3: 100 cm³ of oxygen and 60 cm³ of carbon dioxide. Water is a liquid at r.t.p., so the gas rule does not apply to it.

Checked automatically

(a) Volume of oxygen: 100 cm³ (exact value)
- Common wrong answer 20 : The ratio C₃H₈ : O₂ is 1 : 5, so the oxygen is 5 × 20.
(b) Volume of carbon dioxide: 60 cm³ (exact value)
- Common wrong answer 80 : That is 4 × 20, the water. Carbon dioxide is 3 × 20.
- Common wrong answer 20 : The ratio C₃H₈ : CO₂ is 1 : 3, so the carbon dioxide is 3 × 20.

A full answer includes

(a) 5 × 20 = 100 cm³ of oxygen.
(b) 3 × 20 = 60 cm³ of carbon dioxide.

Also accept

100 cm³; 60 cm³

Watch for

“The water makes 80 cm³ of gas too.” — At r.t.p. water is a liquid, so it has no gas volume in the mole ratio.

10. YT1 · Apply

Model answer. n(N₂) = 72.0 dm³ ÷ 24 dm³/mol = 3.00 mol. Two NaN₃ give three N₂, so the azide is two thirds of the nitrogen: 3.00 × 2/3 = 2.00 mol. M(NaN₃) = 23 + 42 = 65 g/mol, so m = 2.00 × 65 = 130 g.

Checked automatically

Mass of sodium azide: 130 g (3 significant figures)
- Common wrong answer 195 : You used a 1 : 1 ratio. Three N₂ come from two NaN₃: multiply by 2/3.
- Common wrong answer 293 : The ratio is upside down. Moles of NaN₃ = moles of N₂ × 2/3, not × 3/2.

A full answer includes

n(N₂) = 72.0 ÷ 24 = 3.00 mol.
NaN₃ : N₂ = 2 : 3, so n(NaN₃) = 3.00 × 2 ÷ 3 = 2.00 mol.
M(NaN₃) = 23 + 3 × 14 = 65 g/mol; m = 2.00 × 65 = 130 g.

Also accept

130 g

Watch for

“Use the ratio upside down.” — Write the ratio with the substance you want on top: NaN₃ : N₂ = 2 : 3, so moles of NaN₃ = moles of N₂ × 2/3.