A Level H2 Chemistry Stoichiometry Moles Quiz

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A-Level Chemistry H2 Quiz - Stoichiometry Moles

Name: __________________________
Class: __________________________
Date: __________________________
Score: ______ / 50

Duration: 60 minutes
Total Marks: 50

Instructions:

1. Answer all questions.
2. Write your answers in the spaces provided.
3. Show all working for calculation questions. Marks may be awarded for correct steps even if the final answer is incorrect.
4. Use the Data Booklet where relevant.
5. State units where appropriate.

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Section A: Multiple Choice & Short Concepts (10 Marks)

1. Which of the following contains the greatest number of atoms?
A. $1.0 \text{ mol}$ of $\text{H}_2\text{O}$
B. $1.0 \text{ mol}$ of $\text{NH}_3$
C. $1.0 \text{ mol}$ of $\text{CH}_4$
D. $1.0 \text{ mol}$ of $\text{CO}_2$
[1]

2. What is the empirical formula of a compound containing $40.0\%$ carbon, $6.7\%$ hydrogen, and $53.3\%$ oxygen by mass?
A. $\text{CH}_2\text{O}$
B. $\text{C}_2\text{H}_4\text{O}_2$
C. $\text{C}_3\text{H}_6\text{O}_3$
D. $\text{CH}_3\text{O}$
[1]

3. $20.0 \text{ cm}^3$ of a gaseous hydrocarbon $\text{C}_x\text{H}_y$ was exploded with an excess of oxygen. After cooling to room temperature, the volume of the residual gas was $40.0 \text{ cm}^3$ less than the original total volume. When the residual gas was treated with aqueous potassium hydroxide, the volume decreased by a further $40.0 \text{ cm}^3$. What is the molecular formula of the hydrocarbon?
A. $\text{C}_2\text{H}_4$
B. $\text{C}_2\text{H}_6$
C. $\text{C}_3\text{H}_6$
D. $\text{C}_3\text{H}_8$
[1]

4. A sample of hydrated copper(II) sulfate, $\text{CuSO}_4 \cdot x\text{H}_2\text{O}$, has a mass of $2.50 \text{ g}$. Upon heating to constant mass, $1.60 \text{ g}$ of anhydrous $\text{CuSO}_4$ remains. What is the value of $x$?
($A_r$: $\text{Cu} = 63.5, \text{S} = 32.1, \text{O} = 16.0, \text{H} = 1.0$)
A. 3
B. 4
C. 5
D. 6
[1]

5. Which statement about the mole concept is incorrect?
A. One mole of any ideal gas occupies approximately $24.0 \text{ dm}^3$ at room temperature and pressure (r.t.p.).
B. One mole of any substance contains exactly $6.02 \times 10^{23}$ particles.
C. The molar mass of a substance is numerically equal to its relative molecular mass but has units of $\text{g mol}^{-1}$.
D. The number of moles of an element can be calculated by dividing the mass in grams by its relative atomic mass.
[1]

6. Calculate the concentration, in $\text{mol dm}^{-3}$, of a solution prepared by dissolving $5.30 \text{ g}$ of sodium carbonate ($\text{Na}_2\text{CO}_3$) in water to make $250 \text{ cm}^3$ of solution.
($A_r$: $\text{Na} = 23.0, \text{C} = 12.0, \text{O} = 16.0$)
[2]

7. In the reaction between magnesium and hydrochloric acid:
$\text{Mg}(s) + 2\text{HCl}(aq) \rightarrow \text{MgCl}_2(aq) + \text{H}_2(g)$
$0.12 \text{ g}$ of magnesium is reacted with $50.0 \text{ cm}^3$ of $0.20 \text{ mol dm}^{-3}$ hydrochloric acid. Determine the limiting reagent.
($A_r$: $\text{Mg} = 24.3$)
[2]

8. A student prepares a solution by dissolving $2.65 \text{ g}$ of anhydrous sodium carbonate ($\text{Na}_2\text{CO}_3$) in water and making the volume up to $500 \text{ cm}^3$. Calculate the concentration of sodium ions, $[\text{Na}^+]$, in the solution.
($A_r$: $\text{Na} = 23.0, \text{C} = 12.0, \text{O} = 16.0$)
[2]

9. What volume of hydrogen gas, measured at r.t.p., is produced when $0.050 \text{ mol}$ of calcium reacts completely with excess water?
$\text{Ca}(s) + 2\text{H}_2\text{O}(l) \rightarrow \text{Ca(OH)}_2(aq) + \text{H}_2(g)$
(Molar volume of gas at r.t.p. $= 24.0 \text{ dm}^3 \text{ mol}^{-1}$)
[2]

10. $10.0 \text{ cm}^3$ of $0.10 \text{ mol dm}^{-3}$ sulfuric acid is neutralized by $20.0 \text{ cm}^3$ of sodium hydroxide solution. What is the concentration of the sodium hydroxide solution?
$\text{H}_2\text{SO}_4(aq) + 2\text{NaOH}(aq) \rightarrow \text{Na}_2\text{SO}_4(aq) + 2\text{H}_2\text{O}(l)$
[2]

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Section B: Structured Calculations (25 Marks)

11. A mixture of sodium chloride ($\text{NaCl}$) and sodium bromide ($\text{NaBr}$) has a total mass of $2.00 \text{ g}$. The mixture is dissolved in water and treated with excess silver nitrate solution, producing $3.80 \text{ g}$ of a precipitate consisting of silver chloride ($\text{AgCl}$) and silver bromide ($\text{AgBr}$).

(a) Write the ionic equations for the formation of the precipitates.
[2]

(b) Let $x$ be the mass of $\text{NaCl}$ in the original mixture. Derive an expression for the mass of $\text{AgCl}$ formed in terms of $x$.
($A_r$: $\text{Na} = 23.0, \text{Cl} = 35.5, \text{Br} = 79.9, \text{Ag} = 107.9$)
[3]

(c) Calculate the percentage by mass of $\text{NaCl}$ in the original mixture.
[4]

12. Iron(III) oxide reacts with carbon monoxide according to the following equation:
$\text{Fe}_2\text{O}_3(s) + 3\text{CO}(g) \rightarrow 2\text{Fe}(s) + 3\text{CO}_2(g)$

(a) Calculate the maximum mass of iron that can be produced from $10.0 \text{ kg}$ of iron(III) oxide.
($A_r$: $\text{Fe} = 55.8, \text{O} = 16.0$)
[3]

(b) In an industrial process, $6.50 \text{ kg}$ of iron was actually produced. Calculate the percentage yield of the reaction.
[2]

(c) Explain why the actual yield is often lower than the theoretical yield in industrial processes. Give two reasons.
[2]

13. A student performs a titration to determine the concentration of a sulfuric acid solution, $\text{H}_2\text{SO}_4$.
$25.0 \text{ cm}^3$ of $0.150 \text{ mol dm}^{-3}$ sodium hydroxide ($\text{NaOH}$) solution is pipetted into a conical flask. The sulfuric acid is added from a burette. The equation for the reaction is:
$2\text{NaOH}(aq) + \text{H}_2\text{SO}_4(aq) \rightarrow \text{Na}_2\text{SO}_4(aq) + 2\text{H}_2\text{O}(l)$

The following burette readings were recorded:

Titration	Rough	1	2	3
Final reading / $\text{cm}^3$	24.50	23.80	47.90	24.10
Initial reading / $\text{cm}^3$	0.00	0.00	23.80	0.00

(a) Complete the table by calculating the titre for each titration.
[2]

(b) Identify the concordant titres and calculate the mean titre to be used for calculations.
[2]

(c) Calculate the concentration of the sulfuric acid in $\text{mol dm}^{-3}$.
[3]

14. Compound Z is an organic acid containing only carbon, hydrogen, and oxygen. Combustion of $1.00 \text{ g}$ of Z produces $1.47 \text{ g}$ of $\text{CO}_2$ and $0.60 \text{ g}$ of $\text{H}_2\text{O}$.

(a) Calculate the mass of carbon and hydrogen in $1.00 \text{ g}$ of Z.
[2]

(b) Determine the empirical formula of Z.
[3]

(c) The mass spectrum of Z shows a molecular ion peak at $m/z = 60$. Determine the molecular formula of Z.
[2]

15. A mixture of two gases, nitrogen ($\text{N}_2$) and hydrogen ($\text{H}_2$), is placed in a sealed container. The total pressure of the mixture is $100 \text{ kPa}$. The mole fraction of nitrogen is $0.25$.

(a) Calculate the partial pressure of hydrogen in the mixture.
[2]

(b) The gases react to form ammonia ($\text{NH}_3$) according to the equation:
$\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$
If $2.0 \text{ mol}$ of $\text{N}_2$ and $6.0 \text{ mol}$ of $\text{H}_2$ are initially present, and at equilibrium, $1.0 \text{ mol}$ of $\text{NH}_3$ is formed, calculate the total number of moles of gas present at equilibrium.
[3]

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Section C: Advanced Application & Reasoning (15 Marks)

16. An alloy of aluminum and magnesium has a mass of $0.500 \text{ g}$. When reacted with excess hydrochloric acid, it produces $560 \text{ cm}^3$ of hydrogen gas measured at r.t.p. ($1 \text{ mol gas} = 24.0 \text{ dm}^3$ at r.t.p.).

The reactions are:
$2\text{Al}(s) + 6\text{HCl}(aq) \rightarrow 2\text{AlCl}_3(aq) + 3\text{H}_2(g)$
$\text{Mg}(s) + 2\text{HCl}(aq) \rightarrow \text{MgCl}_2(aq) + \text{H}_2(g)$

(a) Calculate the total number of moles of hydrogen gas produced.
[1]

(b) Let $y$ be the mass of aluminum in the alloy. Derive an equation linking $y$ to the total moles of hydrogen produced.
($A_r$: $\text{Al} = 27.0, \text{Mg} = 24.3$)
[3]

(c) Calculate the percentage by mass of magnesium in the alloy.
[2]

17. Hydrated ethanedioic acid (oxalic acid), $\text{H}_2\text{C}_2\text{O}_4 \cdot x\text{H}_2\text{O}$, is used as a primary standard. $1.575 \text{ g}$ of the hydrated acid is dissolved in water and made up to $250 \text{ cm}^3$ in a volumetric flask. $25.0 \text{ cm}^3$ of this solution requires $25.0 \text{ cm}^3$ of $0.100 \text{ mol dm}^{-3}$ sodium hydroxide for neutralization.

$\text{H}_2\text{C}_2\text{O}_4(aq) + 2\text{NaOH}(aq) \rightarrow \text{Na}_2\text{C}_2\text{O}_4(aq) + 2\text{H}_2\text{O}(l)$

(a) Calculate the number of moles of $\text{NaOH}$ used.
[1]

(b) Calculate the number of moles of ethanedioic acid in the $25.0 \text{ cm}^3$ aliquot.
[1]

(c) Calculate the molar mass of the hydrated ethanedioic acid.
[2]

(d) Determine the value of $x$.
($A_r$: $\text{H} = 1.0, \text{C} = 12.0, \text{O} = 16.0$)
[2]

18. A student wishes to prepare $250 \text{ cm}^3$ of a $0.100 \text{ mol dm}^{-3}$ solution of sodium hydroxide ($\text{NaOH}$).

(a) Calculate the mass of solid $\text{NaOH}$ required.
($A_r$: $\text{Na} = 23.0, \text{O} = 16.0, \text{H} = 1.0$)
[2]

(b) Describe the key steps to prepare this standard solution accurately, including the apparatus used.
[3]

19. Calcium carbonate reacts with hydrochloric acid according to the equation:
$\text{CaCO}_3(s) + 2\text{HCl}(aq) \rightarrow \text{CaCl}_2(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g)$
$1.00 \text{ g}$ of calcium carbonate is added to $50.0 \text{ cm}^3$ of $1.00 \text{ mol dm}^{-3}$ hydrochloric acid.

(a) Determine the limiting reagent.
($A_r$: $\text{Ca} = 40.1, \text{C} = 12.0, \text{O} = 16.0$)
[3]

(b) Calculate the volume of carbon dioxide gas produced at r.t.p.
[2]

20. A compound contains $52.2\%$ carbon, $13.0\%$ hydrogen, and $34.8\%$ oxygen by mass.

(a) Determine the empirical formula of the compound.
[3]

(b) If the molar mass of the compound is $46.0 \text{ g mol}^{-1}$, determine its molecular formula.
[2]

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A-Level Chemistry H2 Quiz - Stoichiometry Moles (Answer Key)

1. C
[1]
Reasoning:
A. $\text{H}_2\text{O}$: $3$ atoms/mol $\times 1.0 = 3.0$ mol atoms
B. $\text{NH}_3$: $4$ atoms/mol $\times 1.0 = 4.0$ mol atoms
C. $\text{CH}_4$: $5$ atoms/mol $\times 1.0 = 5.0$ mol atoms
D. $\text{CO}_2$: $3$ atoms/mol $\times 1.0 = 3.0$ mol atoms
$\text{CH}_4$ has the most atoms.

2. A
[1]
Reasoning:
Assume $100 \text{ g}$ 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 = $1 : 2 : 1$. Empirical formula is $\text{CH}_2\text{O}$.

3. A
[1]
Reasoning:
Volume contraction due to reaction and condensation of water.
$\text{C}_x\text{H}_y + (x + y/4)\text{O}_2 \rightarrow x\text{CO}_2 + (y/2)\text{H}_2\text{O}(l)$
Vol $\text{CO}_2$ produced = $40.0 \text{ cm}^3$ (absorbed by KOH).
Since $20.0 \text{ cm}^3$ hydrocarbon produces $40.0 \text{ cm}^3 \text{CO}_2$, $x = 2$.
Contraction = Vol reactants (gas) - Vol products (gas).
Reactants gas: $1$ vol hydrocarbon + $(x+y/4)$ vol $\text{O}_2$.
Products gas: $x$ vol $\text{CO}_2$ (water is liquid).
Change per vol hydrocarbon = $(1 + x + y/4) - x = 1 + y/4$.
Total contraction = $40.0 \text{ cm}^3$ for $20.0 \text{ cm}^3$ hydrocarbon.
Contraction per vol = $40/20 = 2$.
$1 + y/4 = 2 \Rightarrow y/4 = 1 \Rightarrow y = 4$.
Formula is $\text{C}_2\text{H}_4$.

4. C
[1]
Reasoning:
Mass $\text{H}_2\text{O} = 2.50 - 1.60 = 0.90 \text{ g}$.
Moles $\text{H}_2\text{O} = 0.90 / 18.0 = 0.05 \text{ mol}$.
Moles $\text{CuSO}_4 = 1.60 / 159.6 \approx 0.010 \text{ mol}$.
Ratio $\text{CuSO}_4 : \text{H}_2\text{O} = 0.010 : 0.050 = 1 : 5$. $x=5$.

5. B
[1]
Reasoning:
Statement B is incorrect because $6.02 \times 10^{23}$ is an approximation (to 3 significant figures) of the Avogadro constant. The statement claims it contains exactly this number, which is false. The exact number is defined by the Avogadro constant ($N_A \approx 6.022 \times 10^{23}$).
Statement A is accepted in the Singapore syllabus context where molar volume at r.t.p. is approximated as $24.0 \text{ dm}^3$.

6. Calculation
[2]
Working:
Molar mass of $\text{Na}_2\text{CO}_3 = (2 \times 23.0) + 12.0 + (3 \times 16.0) = 106.0 \text{ g mol}^{-1}$.
Moles of $\text{Na}_2\text{CO}_3 = \frac{5.30}{106.0} = 0.050 \text{ mol}$.
Volume $= 250 \text{ cm}^3 = 0.250 \text{ dm}^3$.
Concentration $= \frac{0.050}{0.250} = 0.200 \text{ mol dm}^{-3}$.
Answer: $0.200 \text{ mol dm}^{-3}$

7. Limiting Reagent
[2]
Working:
Moles of $\text{Mg} = \frac{0.12}{24.3} \approx 0.00494 \text{ mol}$.
Moles of $\text{HCl} = 0.20 \times \frac{50.0}{1000} = 0.010 \text{ mol}$.
Stoichiometry: $1 \text{ mol Mg}$ reacts with $2 \text{ mol HCl}$.
Required $\text{HCl}$ for $0.00494 \text{ mol Mg} = 2 \times 0.00494 = 0.00988 \text{ mol}$.
Available $\text{HCl} (0.010) >$ Required $\text{HCl} (0.00988)$.
Therefore, $\text{HCl}$ is in excess and Magnesium is the limiting reagent.
Answer: Magnesium ($\text{Mg}$)

8. Sodium Ion Concentration
[2]
Working:
Molar mass $\text{Na}_2\text{CO}_3 = 106.0 \text{ g mol}^{-1}$.
Moles $\text{Na}_2\text{CO}_3 = \frac{2.65}{106.0} = 0.025 \text{ mol}$.
Volume $= 500 \text{ cm}^3 = 0.500 \text{ dm}^3$.
Concentration of $\text{Na}_2\text{CO}_3 = \frac{0.025}{0.500} = 0.050 \text{ mol dm}^{-3}$.
Each mole of $\text{Na}_2\text{CO}_3$ produces 2 moles of $\text{Na}^+$.
$[\text{Na}^+] = 2 \times 0.050 = 0.100 \text{ mol dm}^{-3}$.
Answer: $0.100 \text{ mol dm}^{-3}$

9. Volume of Hydrogen
[2]
Working:
From equation: $1 \text{ mol Ca}$ produces $1 \text{ mol H}_2$.
Moles $\text{H}_2 = 0.050 \text{ mol}$.
Volume at r.t.p. $= 0.050 \times 24.0 = 1.20 \text{ dm}^3$.
Answer: $1.20 \text{ dm}^3$ (or $1200 \text{ cm}^3$)

10. Concentration of NaOH
[2]
Working:
Moles $\text{H}_2\text{SO}_4 = 0.10 \times \frac{10.0}{1000} = 0.0010 \text{ mol}$.
From equation: $1 \text{ mol H}_2\text{SO}_4$ reacts with $2 \text{ mol NaOH}$.
Moles $\text{NaOH} = 2 \times 0.0010 = 0.0020 \text{ mol}$.
Volume $\text{NaOH} = 20.0 \text{ cm}^3 = 0.020 \text{ dm}^3$.
Concentration $\text{NaOH} = \frac{0.0020}{0.020} = 0.10 \text{ mol dm}^{-3}$.
Answer: $0.10 \text{ mol dm}^{-3}$

11. Mixture Analysis
(a) Ionic Equations [2]
$\text{Ag}^+(aq) + \text{Cl}^-(aq) \rightarrow \text{AgCl}(s)$
$\text{Ag}^+(aq) + \text{Br}^-(aq) \rightarrow \text{AgBr}(s)$

(b) Expression for Mass of AgCl [3]
Molar mass $\text{NaCl} = 23.0 + 35.5 = 58.5 \text{ g mol}^{-1}$.
Molar mass $\text{AgCl} = 107.9 + 35.5 = 143.4 \text{ g mol}^{-1}$.
Moles of $\text{NaCl} = \frac{x}{58.5}$.
From stoichiometry, $1 \text{ mol NaCl} \rightarrow 1 \text{ mol AgCl}$.
Mass of $\text{AgCl} = \text{Moles} \times M_r = \frac{x}{58.5} \times 143.4$.
Expression: $\text{Mass AgCl} = \frac{143.4x}{58.5}$ (or approx $2.451x$)

(c) Percentage by Mass of NaCl [4]
Let mass of $\text{NaCl} = x$. Mass of $\text{NaBr} = 2.00 - x$.
Molar mass $\text{NaBr} = 23.0 + 79.9 = 102.9 \text{ g mol}^{-1}$.
Molar mass $\text{AgBr} = 107.9 + 79.9 = 187.8 \text{ g mol}^{-1}$.
Mass of $\text{AgBr} = \frac{2.00 - x}{102.9} \times 187.8$.
Total precipitate mass $= 3.80 \text{ g}$.
$\frac{143.4x}{58.5} + \frac{187.8(2.00 - x)}{102.9} = 3.80$
$2.451x + 1.825(2.00 - x) = 3.80$
$2.451x + 3.650 - 1.825x = 3.80$
$0.626x = 3.80 - 3.650 = 0.15$
$x = \frac{0.15}{0.626} \approx 0.2396 \text{ g}$.
$\% \text{NaCl} = \frac{0.2396}{2.00} \times 100 = 11.98\%$.
Answer: $12.0\%$ (to 3 s.f.)

12. Iron Production
(a) Max Mass of Iron [3]
Molar mass $\text{Fe}_2\text{O}_3 = (2 \times 55.8) + (3 \times 16.0) = 111.6 + 48.0 = 159.6 \text{ g mol}^{-1}$.
Moles $\text{Fe}_2\text{O}_3 = \frac{10000 \text{ g}}{159.6 \text{ g mol}^{-1}} \approx 62.657 \text{ mol}$.
From equation: $1 \text{ mol Fe}_2\text{O}_3 \rightarrow 2 \text{ mol Fe}$.
Moles $\text{Fe} = 2 \times 62.657 = 125.314 \text{ mol}$.
Mass $\text{Fe} = 125.314 \times 55.8 \approx 6992.5 \text{ g} = 6.99 \text{ kg}$.
Answer: $6.99 \text{ kg}$

(b) Percentage Yield [2]
$\% \text{ Yield} = \frac{\text{Actual}}{\text{Theoretical}} \times 100 = \frac{6.50}{6.9925} \times 100 \approx 92.96\%$.
Answer: $93.0\%$

(c) Reasons for Lower Yield [2]

1. The reaction may be reversible and not go to completion.
2. Side reactions may occur, consuming reactants or producing different products.
   (Other acceptable answers: Loss of product during purification/separation, impure reactants.)

13. Titration Analysis
(a) Table Completion [2]
Titre 1: $23.80 - 0.00 = 23.80$
Titre 2: $47.90 - 23.80 = 24.10$
Titre 3: $24.10 - 0.00 = 24.10$

(b) Mean Titre [2]
Concordant titres are those within $0.10 \text{ cm}^3$ of each other.
Titres 2 and 3 are concordant ($24.10$ and $24.10$). Titre 1 ($23.80$) is not concordant with them.
Mean titre $= \frac{24.10 + 24.10}{2} = 24.10 \text{ cm}^3$.
Answer: $24.10 \text{ cm}^3$

(c) Concentration of Acid [3]
Moles $\text{NaOH} = 0.150 \times \frac{25.0}{1000} = 0.00375 \text{ mol}$.
From equation: $2 \text{ mol NaOH}$ react with $1 \text{ mol H}_2\text{SO}_4$.
Moles $\text{H}_2\text{SO}_4 = \frac{0.00375}{2} = 0.001875 \text{ mol}$.
Volume $\text{H}_2\text{SO}_4 = 24.10 \text{ cm}^3 = 0.02410 \text{ dm}^3$.
Concentration $\text{H}_2\text{SO}_4 = \frac{0.001875}{0.02410} \approx 0.0778 \text{ mol dm}^{-3}$.
Answer: $0.0778 \text{ mol dm}^{-3}$

14. Empirical & Molecular Formula
(a) Mass of C and H [2]
Moles $\text{CO}_2 = \frac{1.47}{44.0} = 0.03341 \text{ mol}$.
Mass $\text{C} = 0.03341 \times 12.0 = 0.4009 \text{ g}$.
Moles $\text{H}_2\text{O} = \frac{0.60}{18.0} = 0.03333 \text{ mol}$.
Moles $\text{H} = 2 \times 0.03333 = 0.06667 \text{ mol}$.
Mass $\text{H} = 0.06667 \times 1.0 = 0.0667 \text{ g}$.
Answer: C: $0.401 \text{ g}$, H: $0.067 \text{ g}$

(b) Empirical Formula [3]
Mass $\text{O} = 1.00 - (0.4009 + 0.0667) = 0.5324 \text{ g}$.
Moles $\text{O} = \frac{0.5324}{16.0} = 0.03328 \text{ mol}$.
Ratio C : H : O
$0.03341 : 0.06667 : 0.03328$
Divide by smallest ($0.03328$):
$1.00 : 2.00 : 1.00$
Empirical Formula: $\text{CH}_2\text{O}$

(c) Molecular Formula [2]
Empirical mass of $\text{CH}_2\text{O} = 12.0 + 2.0 + 16.0 = 30.0$.
$\frac{\text{Molecular Mass}}{\text{Empirical Mass}} = \frac{60}{30} = 2$.
Molecular Formula $= (\text{CH}_2\text{O})_2 = \text{C}_2\text{H}_4\text{O}_2$.
Answer: $\text{C}_2\text{H}_4\text{O}_2$

15. Gas Mixture & Equilibrium
(a) Partial Pressure of Hydrogen [2]
Mole fraction $\text{N}_2 = 0.25$.
Mole fraction $\text{H}_2 = 1 - 0.25 = 0.75$.
Partial Pressure $\text{H}_2 = 0.75 \times 100 \text{ kPa} = 75 \text{ kPa}$.
Answer: $75 \text{ kPa}$

(b) Total Moles at Equilibrium [3]
Initial: $\text{N}_2 = 2.0$, $\text{H}_2 = 6.0$, $\text{NH}_3 = 0$.
Change: To form $1.0 \text{ mol NH}_3$, we need $0.5 \text{ mol N}_2$ and $1.5 \text{ mol H}_2$.
Equilibrium:
$\text{N}_2 = 2.0 - 0.5 = 1.5 \text{ mol}$
$\text{H}_2 = 6.0 - 1.5 = 4.5 \text{ mol}$
$\text{NH}_3 = 1.0 \text{ mol}$
Total moles $= 1.5 + 4.5 + 1.0 = 7.0 \text{ mol}$.
Answer: $7.0 \text{ mol}$

16. Alloy Analysis
(a) Moles of Hydrogen [1]
Volume $= 560 \text{ cm}^3 = 0.560 \text{ dm}^3$.
Moles $\text{H}_2 = \frac{0.560}{24.0} = 0.02333 \text{ mol}$.
Answer: $0.0233 \text{ mol}$

(b) Equation Linking y [3]
Let mass $\text{Al} = y$ g. Then mass $\text{Mg} = (0.500 - y)$ g.
Moles $\text{Al} = \frac{y}{27.0}$. From eq: $2\text{Al} \rightarrow 3\text{H}_2$.
Moles $\text{H}_2$ from $\text{Al} = \frac{3}{2} \times \frac{y}{27.0} = \frac{y}{18.0}$.
Moles $\text{Mg} = \frac{0.500 - y}{24.3}$. From eq: $1\text{Mg} \rightarrow 1\text{H}_2$.
Moles $\text{H}_2$ from $\text{Mg} = \frac{0.500 - y}{24.3}$.
Total Moles $\text{H}_2 = \frac{y}{18.0} + \frac{0.500 - y}{24.3} = 0.02333$.

(c) Percentage Magnesium [2]
Solving the equation:
$\frac{y}{18.0} + \frac{0.500}{24.3} - \frac{y}{24.3} = 0.02333$
$y(\frac{1}{18.0} - \frac{1}{24.3}) = 0.02333 - 0.02058$
$y(0.05556 - 0.04115) = 0.00275$
$y(0.01441) = 0.00275$
$y = \frac{0.00275}{0.01441} \approx 0.191 \text{ g}$ (Mass of Al).
Mass $\text{Mg} = 0.500 - 0.191 = 0.309 \text{ g}$.
$\% \text{Mg} = \frac{0.309}{0.500} \times 100 = 61.8\%$.
Answer: $61.8\%$

17. Hydrated Acid Standard
(a) Moles NaOH [1]
Moles $= 0

% \text{NaCl} = \frac{0.2396}{2.00} \times 100 = 11.98% \approx 12.0%$. *Answer:*$12.0%$

12. Iron Production
(a) Maximum Mass of Iron [3]
Molar mass $\text{Fe}_2\text{O}_3 = (2 \times 55.8) + (3 \times 16.0) = 159.6 \text{ g mol}^{-1}$.
Moles $\text{Fe}_2\text{O}_3 = \frac{10000}{159.6} \approx 62.657 \text{ mol}$.
From equation: $1 \text{ mol Fe}_2\text{O}_3 \rightarrow 2 \text{ mol Fe}$.
Moles $\text{Fe} = 2 \times 62.657 = 125.314 \text{ mol}$.
Mass $\text{Fe} = 125.314 \times 55.8 \approx 6992.5 \text{ g} = 6.99 \text{ kg}$.
Answer: $6.99 \text{ kg}$

(b) Percentage Yield [2]
$\text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100$
$= \frac{6.50}{6.99} \times 100 \approx 92.99\%$.
Answer: $93.0\%$

(c) Reasons for Lower Yield [2]

1. The reaction may be reversible and does not go to completion (equilibrium is established).
2. Side reactions may occur, producing unwanted by-products.
   (Other acceptable answers: Loss of product during purification/separation; Impure reactants.)

13. Titration
(a) Table Completion [2]
Titre = Final Reading - Initial Reading
Rough: $24.50 - 0.00 = 24.50$
1: $23.80 - 0.00 = 23.80$
2: $47.90 - 23.80 = 24.10$
3: $24.10 - 0.00 = 24.10$

Titration	Rough	1	2	3
Titre / $\text{cm}^3$	24.50	23.80	24.10	24.10

(b) Mean Titre [2]
Concordant titres are those within $0.10 \text{ cm}^3$ of each other. Titres 2 and 3 are concordant ($24.10$ and $24.10$). Titre 1 ($23.80$) is not concordant with the others.
Mean titre $= \frac{24.10 + 24.10}{2} = 24.10 \text{ cm}^3$.
Answer: $24.10 \text{ cm}^3$

(c) Concentration of Sulfuric Acid [3]
Moles $\text{NaOH} = 0.150 \times \frac{25.0}{1000} = 0.00375 \text{ mol}$.
From equation: $2 \text{ mol NaOH}$ react with $1 \text{ mol H}_2\text{SO}_4$.
Moles $\text{H}_2\text{SO}_4 = \frac{1}{2} \times 0.00375 = 0.001875 \text{ mol}$.
Volume $\text{H}_2\text{SO}_4 = 24.10 \text{ cm}^3 = 0.02410 \text{ dm}^3$.
Concentration $\text{H}_2\text{SO}_4 = \frac{0.001875}{0.02410} \approx 0.0778 \text{ mol dm}^{-3}$.
Answer: $0.0778 \text{ mol dm}^{-3}$

14. Combustion Analysis
(a) Mass of C and H [2]
Mass C in $\text{CO}_2$: $\frac{12.0}{44.0} \times 1.47 = 0.401 \text{ g}$.
Mass H in $\text{H}_2\text{O}$: $\frac{2.0}{18.0} \times 0.60 = 0.067 \text{ g}$.
Answer: C: $0.401 \text{ g}$, H: $0.067 \text{ g}$

(b) Empirical Formula [3]
Mass O $= 1.00 - (0.401 + 0.067) = 0.532 \text{ g}$.
Moles C $= \frac{0.401}{12.0} = 0.0334$.
Moles H $= \frac{0.067}{1.0} = 0.067$.
Moles O $= \frac{0.532}{16.0} = 0.0333$.
Ratio C:H:O $\approx 0.0334 : 0.067 : 0.0333 \approx 1 : 2 : 1$.
Empirical Formula: $\text{CH}_2\text{O}$.

(c) Molecular Formula [2]
Empirical mass of $\text{CH}_2\text{O} = 12.0 + 2.0 + 16.0 = 30.0$.
$\frac{\text{Molecular Mass}}{\text{Empirical Mass}} = \frac{60}{30} = 2$.
Molecular Formula $= (\text{CH}_2\text{O})_2 = \text{C}_2\text{H}_4\text{O}_2$.
Answer: $\text{C}_2\text{H}_4\text{O}_2$

15. Gas Mixtures
(a) Partial Pressure of Hydrogen [2]
Mole fraction $\text{N}_2 = 0.25$.
Mole fraction $\text{H}_2 = 1 - 0.25 = 0.75$.
Partial Pressure $\text{H}_2 = 0.75 \times 100 \text{ kPa} = 75 \text{ kPa}$.
Answer: $75 \text{ kPa}$

(b) Total Moles at Equilibrium [3]
Equation: $\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3$
Initial: $2.0 \quad 6.0 \quad 0$
Change: $-0.5 \quad -1.5 \quad +1.0$ (Since $1.0 \text{ mol NH}_3$ formed, ratio is $1:3:2$)
Equilibrium: $1.5 \quad 4.5 \quad 1.0$
Total moles $= 1.5 + 4.5 + 1.0 = 7.0 \text{ mol}$.
Answer: $7.0 \text{ mol}$

16. Alloy Analysis
(a) Moles of Hydrogen [1]
Volume $= 560 \text{ cm}^3 = 0.560 \text{ dm}^3$.
Moles $\text{H}_2 = \frac{0.560}{24.0} \approx 0.02333 \text{ mol}$.
Answer: $0.0233 \text{ mol}$

(b) Equation Linking $y$ [3]
Let mass $\text{Al} = y \text{ g}$. Then mass $\text{Mg} = (0.500 - y) \text{ g}$.
Moles $\text{Al} = \frac{y}{27.0}$. From $2\text{Al} \rightarrow 3\text{H}_2$, moles $\text{H}_2$ from Al $= \frac{3}{2} \times \frac{y}{27.0} = \frac{y}{18.0}$.
Moles $\text{Mg} = \frac{0.500 - y}{24.3}$. From $\text{Mg} \rightarrow \text{H}_2$, moles $\text{H}_2$ from Mg $= \frac{0.500 - y}{24.3}$.
Total moles $\text{H}_2 = \frac{y}{18.0} + \frac{0.500 - y}{24.3} = 0.02333$.

(c) Percentage by Mass of Magnesium [2]
Solving the equation:
$\frac{y}{18.0} + \frac{0.500 - y}{24.3} = 0.02333$
Multiply by $18.0 \times 24.3 = 437.4$:
$24.3y + 18.0(0.500 - y) = 0.02333 \times 437.4$
$24.3y + 9.0 - 18.0y = 10.205$
$6.3y = 1.205$
$y \approx 0.1913 \text{ g}$ (Mass of Al).
Mass of Mg $= 0.500 - 0.1913 = 0.3087 \text{ g}$.
$\% \text{Mg} = \frac{0.3087}{0.500} \times 100 \approx 61.7\%$.
Answer: $61.7\%$

17. Hydrated Acid Standard
(a) Moles of NaOH [1]
Moles $\text{NaOH} = 0.100 \times \frac{25.0}{1000} = 0.00250 \text{ mol}$.
Answer: $0.00250 \text{ mol}$

(b) Moles of Acid in Aliquot [1]
Ratio $\text{Acid}:\text{NaOH} = 1:2$.
Moles Acid $= \frac{1}{2} \times 0.00250 = 0.00125 \text{ mol}$.
Answer: $0.00125 \text{ mol}$

(c) Molar Mass of Hydrated Acid [2]
Moles in $250 \text{ cm}^3$ flask $= 0.00125 \times \frac{250}{25} = 0.0125 \text{ mol}$.
Molar Mass $= \frac{\text{Mass}}{\text{Moles}} = \frac{1.575}{0.0125} = 126.0 \text{ g mol}^{-1}$.
Answer: $126.0 \text{ g mol}^{-1}$

(d) Value of $x$ [2]
Molar mass of anhydrous $\text{H}_2\text{C}_2\text{O}_4 = (2 \times 1.0) + (2 \times 12.0) + (4 \times 16.0) = 90.0 \text{ g mol}^{-1}$.
Mass of water in formula $= 126.0 - 90.0 = 36.0 \text{ g mol}^{-1}$.
Molar mass $\text{H}_2\text{O} = 18.0 \text{ g mol}^{-1}$.
$x = \frac{36.0}{18.0} = 2$.
Answer: $x = 2$

18. Solution Preparation
(a) Mass of NaOH [2]
Moles required $= 0.100 \times \frac{250}{1000} = 0.0250 \text{ mol}$.
Molar mass $\text{NaOH} = 23.0 + 16.0 + 1.0 = 40.0 \text{ g mol}^{-1}$.
Mass $= 0.0250 \times 40.0 = 1.00 \text{ g}$.
Answer: $1.00 \text{ g}$

(b) Key Steps [3]

1. Weigh $1.00 \text{ g}$ of solid $\text{NaOH}$ accurately using a balance (in a beaker/watch glass as it is hygroscopic).
2. Dissolve the solid in a small amount of distilled water in a beaker. Stir until fully dissolved.
3. Transfer the solution quantitatively to a $250 \text{ cm}^3$ volumetric flask using a funnel. Rinse the beaker and funnel with distilled water and add washings to the flask.
4. Add distilled water to the flask until the meniscus reaches the graduation mark. Stopper and invert to mix.

19. Limiting Reagent & Gas Volume
(a) Limiting Reagent [3]
Moles $\text{CaCO}_3 = \frac{1.00}{100.1} \approx 0.00999 \text{ mol}$.
Moles $\text{HCl} = 1.00 \times \frac{50.0}{1000} = 0.0500 \text{ mol}$.
Stoichiometry: $1 \text{ mol CaCO}_3$ reacts with $2 \text{ mol HCl}$.
Required $\text{HCl}$ for $0.00999 \text{ mol CaCO}_3 = 2 \times 0.00999 = 0.01998 \text{ mol}$.
Available $\text{HCl} (0.0500) >$ Required $\text{HCl} (0.01998)$.
Therefore, $\text{HCl}$ is in excess.
Answer: Calcium Carbonate ($\text{CaCO}_3$) is the limiting reagent.

(b) Volume of $\text{CO}_2$ [2]
From equation: $1 \text{ mol CaCO}_3 \rightarrow 1 \text{ mol CO}_2$.
Moles $\text{CO}_2 = 0.00999 \text{ mol}$.
Volume at r.t.p. $= 0.00999 \times 24.0 \approx 0.240 \text{ dm}^3$.
Answer: $0.240 \text{ dm}^3$ (or $240 \text{ cm}^3$)

20. Empirical and Molecular Formula
(a) Empirical Formula [3]
Assume $100 \text{ g}$.
C: $52.2/12.0 = 4.35$ mol
H: $13.0/1.0 = 13.0$ mol
O: $34.8/16.0 = 2.175$ mol
Divide by smallest ($2.175$):
C: $4.35/2.175 = 2$
H: $13.0/2.175 \approx 5.98 \approx 6$
O: $2.175/2.175 = 1$
Empirical Formula: $\text{C}_2\text{H}_6\text{O}$.

(b) Molecular Formula [2]
Empirical mass $\text{C}_2\text{H}_6\text{O} = (2 \times 12.0) + (6 \times 1.0) + 16.0 = 46.0 \text{ g mol}^{-1}$.
Given molar mass $= 46.0 \text{ g mol}^{-1}$.
Ratio $= 1$.
Molecular Formula: $\text{C}_2\text{H}_6\text{O}$.