Answers · Concentration, gas volumes and yield
Concentration, gas volumes and yield — final check key
Check meaning, not exact wording. These are teaching points, not official Cambridge marks.
1. YK1 · Final check
Model answer. One mole of any gas takes up 24 dm³ = 24 000 cm³ at r.t.p. (a) 0.150 × 24 = 3.60 dm³. (b) 0.0100 × 24 000 = 240 cm³.
Checked automatically
- (a) Volume of nitrogen: 3.60 dm³ (3 significant figures)
- (b) Volume of chlorine: 240 cm³ (3 significant figures)
A full answer includes
- (a) 0.150 × 24 = 3.60 dm³.
- (b) 0.0100 × 24 000 = 240 cm³.
Also accept
- 3.60 dm³; 240 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.
2. YK2 · Final check
Model answer. M(Na2CO3) = 2 × 23 + 12 + 3 × 16 = 106 g/mol, so n = 1.06 ÷ 106 = 0.0100 mol. One Na2CO3 gives one CO2, so 0.0100 mol of gas: 0.0100 × 24 000 = 240 cm³.
Checked automatically
- Volume of carbon dioxide: 240 cm³ (3 significant figures)
A full answer includes
- M(Na2CO3) = 106 g/mol; n = 1.06 ÷ 106 = 0.0100 mol.
- Na2CO3 : CO2 = 1 : 1, so 0.0100 mol of CO2.
- V = 0.0100 × 24 000 = 240 cm³.
Also accept
- 240 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.
3. YK3 · Final check
Model answer. n(Al) = 5.40 ÷ 27 = 0.200 mol. Four Al make two Al2O3, so the oxide is half: 0.100 mol. M(Al2O3) = 2 × 27 + 3 × 16 = 102 g/mol, so m = 0.100 × 102 = 10.2 g.
Checked automatically
- Mass of aluminium oxide: 10.2 g (3 significant figures)
A full answer includes
- n(Al) = 5.40 ÷ 27 = 0.200 mol.
- Al : Al2O3 = 4 : 2, so 0.100 mol of Al2O3.
- M(Al2O3) = 102 g/mol; m = 0.100 × 102 = 10.2 g.
Also accept
- 10.2 g
Watch for
- “The masses react in the ratio of the numbers in the equation.” — The numbers in an equation count particles, so they give the ratio of moles. Change masses to moles first.
4. YK4 · Final check
Model answer. n(Zn) = 0.0100 mol and n(HCl) = 0.0250 mol. Dividing by the numbers in the equation gives 0.0100 for zinc and 0.0125 for the acid, so zinc is limiting. One Zn gives one H2: 0.0100 × 24 000 = 240 cm³.
Checked automatically
- (b) Volume of hydrogen: 240 cm³ (3 significant figures)
A full answer includes
- n(Zn) = 0.650 ÷ 65 = 0.0100 mol; n(HCl) = 0.500 × 0.0500 = 0.0250 mol.
- (a) Zn 0.0100 ÷ 1 = 0.0100; HCl 0.0250 ÷ 2 = 0.0125: zinc is limiting.
- (b) 0.0100 mol of H2 × 24 000 = 240 cm³.
Also accept
- (a) zinc (b) 240 cm³
Watch for
- “The limiting reactant is the one with fewer moles.” — Divide each amount by its number in the equation first. In Mg + 2HCl, the acid can run out even when it has more moles, because each Mg needs two HCl.
5. YK5 · Final check
Model answer. Each H2SO4 needs two NaOH, so 0.00500 mol of sodium hydroxide. V = n ÷ c = 0.00500 ÷ 0.100 = 0.0500 dm³, which is 50.0 cm³.
Checked automatically
- Volume of sodium hydroxide solution: 50.0 cm³ (3 significant figures)
A full answer includes
- H2SO4 : NaOH = 1 : 2, so 0.00500 mol of NaOH.
- V = 0.00500 ÷ 0.100 = 0.0500 dm³ = 50.0 cm³.
Also accept
- 50.0 cm³
Watch for
- “Use the mole ratio on the volumes of solutions.” — Change to moles, use the equation, then V = n ÷ c.
6. YK6 · Final check
Model answer. n(Mg) = 0.100 mol; 2Mg give 2MgO, so 0.100 mol of MgO: 4.00 g. Percentage yield = 3.60 ÷ 4.00 × 100 = 90.0 %.
Checked automatically
- (a) Theoretical yield: 4.00 g (3 significant figures)
- (b) Percentage yield: 90.0 % (3 significant figures)
A full answer includes
- (a) n(Mg) = 2.40 ÷ 24 = 0.100 mol, so 0.100 mol of MgO: 0.100 × 40 = 4.00 g.
- (b) 3.60 ÷ 4.00 × 100 = 90.0 %.
Also accept
- 4.00 g; 90.0 %
Watch for
- “Percentage yield = theoretical yield ÷ actual yield × 100.” — Put the mass you actually collected on top: actual ÷ theoretical × 100. A dry, pure product never gives more than 100 %.
7. YK7 · Final check
Model answer. Urea has Mr = 60 and two nitrogen atoms, 28 of the 60: 46.7 % nitrogen. The sample is 1.90 g of urea in 2.00 g: 95.0 % pure.
Checked automatically
- (a) Nitrogen in urea: 46.7 % (3 significant figures)
- (b) Percentage purity: 95.0 % (3 significant figures)
A full answer includes
- (a) Mr = 12 + 16 + 2 × (14 + 2) = 60; nitrogen 2 × 14 = 28; 28 ÷ 60 × 100 = 46.7 %.
- (b) 1.90 ÷ 2.00 × 100 = 95.0 %.
Also accept
- 46.7 %; 95.0 %
Watch for
- “Use one nitrogen atom.” — The bracket doubles everything inside it: CO(NH2)2 has two nitrogen atoms.
8. YK8 · Final check
Model answer. In 100 g: carbon 3.33 mol, hydrogen 6.7 mol and oxygen 3.33 mol, a ratio of 1 : 2 : 1. The empirical formula is CH2O, with mass 30. 180 ÷ 30 = 6, so the molecular formula is C6H12O6.
Checked automatically
- (a) Mass of one empirical formula: 30 (exact value)
- (b) Empirical-formula units in one molecule: 6 (exact value)
A full answer includes
- In 100 g: C 40.0 ÷ 12 = 3.33, H 6.7 ÷ 1 = 6.7, O 53.3 ÷ 16 = 3.33 mol.
- C : H : O = 1 : 2 : 1, so CH2O, mass 12 + 2 + 16 = 30.
- 180 ÷ 30 = 6, so C6H12O6.
Also accept
- 30; 6: CH2O, C6H12O6
Watch for
- “Round a ratio such as 1 : 1.5 to the nearest whole number.” — Only round a number very close to a whole one, such as 2.98. Multiply a ratio ending in .5 by 2, and one ending in .33 or .67 by 3.