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A Level H1 Chemistry Stoichiometry Moles Quiz

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Questions

A-Level Chemistry H1 Quiz - Stoichiometry Moles

Name: __________________________
Class: __________________________
Date: __________________________
Score: ________ / 45

Duration: 45 minutes
Total Marks: 45

Instructions:

Answer all questions.
Write your answers in the spaces provided.
Show all working clearly. Marks may be awarded for correct working even if the final answer is incorrect.
Use the Data Booklet where appropriate.
Give numerical answers to 3 significant figures unless otherwise stated.

Section A: Basic Mole Calculations and Formulae (Questions 1–5)

1. Calculate the number of moles of atoms present in 12.0 g of magnesium.
[Ar: Mg = 24.3]
[1]

2. A sample of gas occupies 480 cm³ at room temperature and pressure (r.t.p.). Calculate the number of moles of gas present.
[Molar volume of gas at r.t.p. = 24.0 dm³ mol⁻¹]
[1]

3. Determine the empirical formula of a compound containing 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass.
[Ar: C = 12.0, H = 1.0, O = 16.0]
[2]

4. The empirical formula of a hydrocarbon is CH₂. Its relative molecular mass is 56.0. Determine its molecular formula.
[Ar: C = 12.0, H = 1.0]
[1]

5. Calculate the mass of 0.050 mol of sodium hydroxide (NaOH).
[Ar: Na = 23.0, O = 16.0, H = 1.0]
[1]

Section B: Solutions and Concentrations (Questions 6–10)

6. 5.85 g of sodium chloride (NaCl) is dissolved in water and the solution is made up to 500 cm³ in a volumetric flask. Calculate the concentration of the solution in mol dm⁻³.
[Ar: Na = 23.0, Cl = 35.5]
[2]

7. Calculate the volume of 2.0 mol dm⁻³ hydrochloric acid required to prepare 250 cm³ of 0.50 mol dm⁻³ hydrochloric acid by dilution.
[2]

8. 25.0 cm³ of 0.100 mol dm⁻³ sulfuric acid (H₂SO₄) is neutralized by sodium hydroxide (NaOH) solution.
The equation for the reaction is:
$H_2SO_4(aq) + 2NaOH(aq) \rightarrow Na_2SO_4(aq) + 2H_2O(l)$
Calculate the amount (in moles) of NaOH required for complete neutralization.
[2]

9. In a titration, 20.0 cm³ of an unknown concentration of potassium hydroxide (KOH) solution required 25.0 cm³ of 0.100 mol dm⁻³ nitric acid (HNO₃) for neutralization.
The equation is:
$KOH(aq) + HNO_3(aq) \rightarrow KNO_3(aq) + H_2O(l)$
Calculate the concentration of the KOH solution in mol dm⁻³.
[2]

10. A student mixes 100 cm³ of 0.20 mol dm⁻³ NaCl solution with 100 cm³ of 0.20 mol dm⁻³ KCl solution. Assuming volumes are additive, calculate the final concentration of chloride ions ( $Cl^-$ ) in the mixture.
[2]

Section C: Gas Laws and Stoichiometry (Questions 11–15)

11. Calculate the volume occupied by 0.25 mol of oxygen gas at 298 K and 100 kPa.
[Gas constant $R = 8.31 \text{ J K}^{-1} \text{ mol}^{-1}$ ]
[2]

12. 1.00 g of calcium carbonate ( $CaCO_3$ ) is heated strongly until decomposition is complete.
$CaCO_3(s) \rightarrow CaO(s) + CO_2(g)$
Calculate the volume of carbon dioxide gas produced at r.t.p.
[Ar: Ca = 40.1, C = 12.0, O = 16.0; Molar volume at r.t.p. = 24.0 dm³ mol⁻¹]
[3]

13. Nitrogen and hydrogen react to form ammonia according to the equation:
$N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$
If 100 cm³ of nitrogen is mixed with 200 cm³ of hydrogen and allowed to react completely (all volumes at same T and P), calculate the volume of ammonia produced.
[2]

14. A hydrocarbon $C_xH_y$ undergoes complete combustion.
$C_xH_y + (x + \frac{y}{4})O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O$
10 cm³ of the hydrocarbon reacts with 50 cm³ of oxygen to produce 30 cm³ of carbon dioxide. All volumes are measured at the same temperature and pressure. Determine the values of $x$ and $y$ .
[3]

15. 0.150 g of a volatile liquid is vaporized in a gas syringe at 100°C and 101 kPa. The volume of the vapor is 65.0 cm³. Calculate the relative molecular mass ( $M_r$ ) of the liquid.
[ $R = 8.31 \text{ J K}^{-1} \text{ mol}^{-1}$ ]
[3]

Section D: Advanced Stoichiometry and Yield (Questions 16–20)

16. Copper(II) oxide reacts with dilute sulfuric acid to form copper(II) sulfate and water.
$CuO(s) + H_2SO_4(aq) \rightarrow CuSO_4(aq) + H_2O(l)$
In an experiment, 4.00 g of CuO is added to 50.0 cm³ of 1.00 mol dm⁻³ $H_2SO_4$ .
[Ar: Cu = 63.5, O = 16.0]
(a) Determine the limiting reagent.
[2]

(b) Calculate the maximum mass of copper(II) sulfate ( $CuSO_4$ ) that can be formed.
[Ar: S = 32.1]
[2]

17. In the reaction described in Question 16, the actual mass of $CuSO_4 \cdot 5H_2O$ crystals obtained after crystallization was 8.50 g. Calculate the percentage yield of the hydrated salt.
[Ar: H = 1.0, O = 16.0, S = 32.1, Cu = 63.5]
[3]

18. A mixture of sodium carbonate ( $Na_2CO_3$ ) and sodium chloride ( $NaCl$ ) has a total mass of 2.50 g. The mixture is dissolved in water and titrated with 0.500 mol dm⁻³ HCl. 20.0 cm³ of the acid is required for complete reaction.
$Na_2CO_3(aq) + 2HCl(aq) \rightarrow 2NaCl(aq) + H_2O(l) + CO_2(g)$
Calculate the percentage by mass of sodium carbonate in the original mixture.
[Ar: Na = 23.0, C = 12.0, O = 16.0]
[4]

19. Iron(III) oxide reacts with carbon monoxide to produce iron and carbon dioxide.
$Fe_2O_3(s) + 3CO(g) \rightarrow 2Fe(s) + 3CO_2(g)$
Calculate the mass of iron produced when 1.00 kg of $Fe_2O_3$ is reduced by excess CO.
[Ar: Fe = 55.8, O = 16.0]
[3]

20. A hydrated salt $MgSO_4 \cdot xH_2O$ is heated to constant mass. 2.46 g of the hydrated salt yields 1.20 g of anhydrous $MgSO_4$ . Calculate the value of $x$ .
[Ar: Mg = 24.3, S = 32.1, O = 16.0, H = 1.0]
[3]

Answers

A-Level Chemistry H1 Quiz - Stoichiometry Moles (Answer Key)

1.
$n = \frac{m}{M_r} = \frac{12.0}{24.3} = 0.494$ mol
Answer: 0.494 mol [1]

2.
Volume in dm³ = $480 / 1000 = 0.480$ dm³
$n = \frac{V}{V_m} = \frac{0.480}{24.0} = 0.0200$ mol
Answer: 0.0200 mol [1]

3.
Assume 100 g sample.
C: $40.0 / 12.0 = 3.33$ mol
H: $6.7 / 1.0 = 6.7$ mol
O: $53.3 / 16.0 = 3.33$ mol
Ratio C:H:O = $3.33 : 6.7 : 3.33 \approx 1 : 2 : 1$
Answer: $CH_2O$ [2]

4.
Empirical mass of $CH_2 = 12.0 + 2(1.0) = 14.0$
Ratio = $56.0 / 14.0 = 4$
Molecular formula = $C_4H_8$
Answer: $C_4H_8$ [1]

5.
$M_r(NaOH) = 23.0 + 16.0 + 1.0 = 40.0$
$m = n \times M_r = 0.050 \times 40.0 = 2.0$ g
Answer: 2.0 g [1]

6.
$M_r(NaCl) = 23.0 + 35.5 = 58.5$
$n(NaCl) = 5.85 / 58.5 = 0.100$ mol
Volume = $500 \text{ cm}^3 = 0.500 \text{ dm}^3$
Concentration = $0.100 / 0.500 = 0.200$ mol dm⁻³
Answer: 0.200 mol dm⁻³ [2]

7.
Using $C_1V_1 = C_2V_2$
$2.0 \times V_1 = 0.50 \times 250$
$V_1 = \frac{125}{2.0} = 62.5$ cm³
Answer: 62.5 cm³ [2]

8.
$n(H_2SO_4) = 0.100 \times \frac{25.0}{1000} = 0.00250$ mol
From equation, ratio $H_2SO_4 : NaOH = 1 : 2$
$n(NaOH) = 2 \times 0.00250 = 0.00500$ mol
Answer: 0.00500 mol [2]

9.
$n(HNO_3) = 0.100 \times \frac{25.0}{1000} = 0.00250$ mol
Ratio $KOH : HNO_3 = 1 : 1$
$n(KOH) = 0.00250$ mol
$C(KOH) = \frac{0.00250}{20.0/1000} = \frac{0.00250}{0.020} = 0.125$ mol dm⁻³
Answer: 0.125 mol dm⁻³ [2]

10.
Total volume = $100 + 100 = 200$ cm³ = 0.200 dm³
Moles $Cl^-$ from NaCl = $0.100 \times 0.20 = 0.020$ mol
Moles $Cl^-$ from KCl = $0.100 \times 0.20 = 0.020$ mol
Total moles $Cl^- = 0.040$ mol
$[Cl^-] = \frac{0.040}{0.200} = 0.20$ mol dm⁻³
Answer: 0.20 mol dm⁻³ [2]

11.
$T = 298$ K, $P = 100,000$ Pa, $n = 0.25$ mol
$V = \frac{nRT}{P} = \frac{0.25 \times 8.31 \times 298}{100,000}$
$V = \frac{619.095}{100,000} = 0.00619$ m³
Convert to dm³: $0.00619 \times 1000 = 6.19$ dm³
Answer: 6.19 dm³ [2]

12.
$M_r(CaCO_3) = 40.1 + 12.0 + 3(16.0) = 100.1$
$n(CaCO_3) = 1.00 / 100.1 = 0.00999$ mol
Ratio $CaCO_3 : CO_2 = 1 : 1$
$n(CO_2) = 0.00999$ mol
$V(CO_2) = 0.00999 \times 24.0 = 0.23976$ dm³
Answer: 0.240 dm³ (or 240 cm³) [3]

13.
Ratio $N_2 : H_2 = 1 : 3$ .
100 cm³ $N_2$ requires 300 cm³ $H_2$ .
Only 200 cm³ $H_2$ available, so $H_2$ is limiting.
Ratio $H_2 : NH_3 = 3 : 2$ .
Vol $NH_3 = \frac{2}{3} \times 200 = 133$ cm³
Answer: 133 cm³ [2]

14.
Ratio Hydrocarbon : $O_2$ : $CO_2$ = $10 : 50 : 30 = 1 : 5 : 3$ .
From equation: 1 mol hydrocarbon produces $x$ mol $CO_2$ .
So $x = 3$ .
Oxygen balance: $x + y/4 = 5 \Rightarrow 3 + y/4 = 5 \Rightarrow y/4 = 2 \Rightarrow y = 8$ .
Answer: $x = 3, y = 8$ ( $C_3H_8$ ) [3]

15.
$m = 0.150$ g, $T = 100 + 273 = 373$ K, $P = 101,000$ Pa, $V = 65.0 \times 10^{-6}$ m³
$n = \frac{PV}{RT} = \frac{101,000 \times 65.0 \times 10^{-6}}{8.31 \times 373} = \frac{6.565}{3099.63} = 0.002118$ mol
$M_r = \frac{m}{n} = \frac{0.150}{0.002118} = 70.8$
Answer: 70.8 [3]

16.
(a)
$n(CuO) = 4.00 / (63.5 + 16.0) = 4.00 / 79.5 = 0.0503$ mol
$n(H_2SO_4) = 1.00 \times 0.050 = 0.0500$ mol
Ratio is 1:1. Since $0.0500 < 0.0503$ , $H_2SO_4$ is limiting.
Answer: Sulfuric acid ( $H_2SO_4$ ) [2]

(b)
Limiting reagent is $H_2SO_4$ (0.0500 mol).
Ratio $H_2SO_4 : CuSO_4 = 1 : 1$ .
$n(CuSO_4) = 0.0500$ mol.
$M_r(CuSO_4) = 63.5 + 32.1 + 4(16.0) = 159.6$
Mass = $0.0500 \times 159.6 = 7.98$ g
Answer: 7.98 g [2]

17.
Theoretical moles of $CuSO_4 = 0.0500$ mol.
Hydrated salt is $CuSO_4 \cdot 5H_2O$ .
$M_r = 159.6 + 5(18.0) = 249.6$
Theoretical mass = $0.0500 \times 249.6 = 12.48$ g
% Yield = $\frac{8.50}{12.48} \times 100 = 68.1\%$
Answer: 68.1% [3]

18.
$n(HCl) = 0.500 \times \frac{20.0}{1000} = 0.0100$ mol
Ratio $Na_2CO_3 : HCl = 1 : 2$ .
$n(Na_2CO_3) = 0.0100 / 2 = 0.00500$ mol
$M_r(Na_2CO_3) = 2(23.0) + 12.0 + 3(16.0) = 106.0$
Mass $Na_2CO_3 = 0.00500 \times 106.0 = 0.530$ g
% Mass = $\frac{0.530}{2.50} \times 100 = 21.2\%$
Answer: 21.2% [4]

19.
$M_r(Fe_2O_3) = 2(55.8) + 3(16.0) = 159.6$
Mass = 1000 g
$n(Fe_2O_3) = 1000 / 159.6 = 6.266$ mol
Ratio $Fe_2O_3 : Fe = 1 : 2$ .
$n(Fe) = 2 \times 6.266 = 12.53$ mol
Mass $Fe = 12.53 \times 55.8 = 699$ g
Answer: 699 g (or 0.699 kg) [3]

20.
Mass water lost = $2.46 - 1.20 = 1.26$ g
$n(H_2O) = 1.26 / 18.0 = 0.0700$ mol
$n(MgSO_4) = 1.20 / (24.3 + 32.1 + 64.0) = 1.20 / 120.4 = 0.00997$ mol
Ratio $x = \frac{n(H_2O)}{n(MgSO_4)} = \frac{0.0700}{0.00997} \approx 7.02$
Answer: $x = 7$ [3]