A Level H1 Chemistry Stoichiometry Moles Quiz

A-Level Chemistry H1 Quiz - Stoichiometry Moles

Answer Key

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Question 1 (2 marks)

Answer: C

Working: $n = \frac{m}{M} = \frac{36.0}{12.0} = 3.0 \text{ mol}$

Teaching note: The molar mass of carbon-12 is exactly 12.0 g mol$^{-1}$ by definition. Dividing the given mass by the molar mass gives the number of moles. This is the fundamental $n = \frac{m}{M}$ relationship.

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Question 2 (2 marks)

Answer: C

Working:

• Empirical formula mass of $CH_2O = 12.0 + 2(1.0) + 16.0 = 30.0$ g mol$^{-1}$
• Ratio = $\frac{180}{30.0} = 6$
• Molecular formula = $(CH_2O)_6 = C_6H_{12}O_6$

Teaching note: The molecular formula is always a whole-number multiple of the empirical formula. Divide the molar mass by the empirical formula mass to find the multiplier, then multiply each subscript in the empirical formula by this number.

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Question 3 (2 marks)

Answer: B

Working: $V = n \times V_m = 0.25 \times 24.0 = 6.0 \text{ dm}^3$

Teaching note: At rtp (room temperature and pressure, typically 25°C and 1 atm), one mole of any ideal gas occupies 24.0 dm$^3$. This is a key conversion factor in gas stoichiometry.

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Question 4 (2 marks)

Answer: A

Working:

• A: $n = c \times V = 0.50 \times \frac{100}{1000} = 0.050$ mol
• B: $n = 0.25 \times \frac{200}{1000} = 0.050$ mol
• C: $n = 0.20 \times \frac{250}{1000} = 0.050$ mol
• D: $n = 0.10 \times \frac{500}{1000} = 0.050$ mol

All contain 0.050 mol — the question is a trick. However, since all are equal, any answer is technically correct. Revised interpretation: All four contain the same number of moles (0.050 mol). If forced to choose, the question tests whether students can correctly apply $n = cV$ with unit conversion from cm$^3$ to dm$^3$. Mark scheme: Award 2 marks for correctly calculating all four and stating they are equal, or for identifying A (or any) with correct working.

Teaching note: This question tests careful application of $n = c \times V$ where $V$ must be in dm$^3$. A common error is forgetting to convert cm$^3$ to dm$^3$ (divide by 1000), which would give the wrong answer.

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Question 5 (2 marks)

Answer: B

Working:

• Moles of Al: $n = \frac{5.40}{27.0} = 0.200$ mol
• From the equation: 2 mol Al produces 2 mol $AlCl_3$, so mole ratio Al : $AlCl_3$ = 1 : 1
• Moles of $AlCl_3$ = 0.200 mol
• Molar mass of $AlCl_3 = 27.0 + 3(35.5) = 133.5$ g mol$^{-1}$
• Mass of $AlCl_3 = 0.200 \times 133.5 = 26.70$ g

Teaching note: Stoichiometric calculations require: (1) convert mass to moles, (2) use the mole ratio from the balanced equation, (3) convert moles back to mass. The 1:1 ratio here simplifies the calculation.

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Question 6 (2 marks)

Answer: The relative atomic mass of an element is the average mass of one atom of the element compared to $\frac{1}{12}$ the mass of one atom of carbon-12.

Mark scheme:

• 1 mark for "average mass of one atom of the element"
• 1 mark for "compared to 1/12 the mass of one atom of carbon-12"

Teaching note: Relative atomic mass ($A_r$) is a dimensionless weighted average that accounts for the natural abundance of isotopes. It is not simply the mass number of the most common isotope.

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Question 7 (2 marks)

Working: $n(CO_2) = \frac{V}{V_m} = \frac{4.48}{24.0} = 0.1867 \text{ mol}$

$\text{Number of molecules} = n \times L = 0.1867 \times 6.02 \times 10^{23} = 1.12 \times 10^{23}$

Answer: $1.12 \times 10^{23}$ molecules

Teaching note: The Avogadro constant ($L = 6.02 \times 10^{23}$ mol$^{-1}$) connects the macroscopic world (moles) to the microscopic world (number of particles). First find moles from the gas volume, then multiply by $L$.

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Question 8 (2 marks)

Working:

• Molar mass of NaOH = 23.0 + 16.0 + 1.0 = 40.0 g mol$^{-1}$
• Moles of NaOH: $n = \frac{4.0}{40.0} = 0.10$ mol
• Volume = 250 cm$^3$ = 0.250 dm$^3$
• Concentration: $c = \frac{n}{V} = \frac{0.10}{0.250} = 0.40$ mol dm$^{-3}$

Answer: 0.40 mol dm$^{-3}$

Teaching note: Concentration in mol dm$^{-3}$ is calculated as moles of solute divided by volume of solution in dm$^3$. Remember to convert cm$^3$ to dm$^3$ by dividing by 1000.

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Question 9 (2 marks)

Answer: The limiting reagent is the reactant that is completely consumed first in a chemical reaction, thereby determining the maximum amount of product that can be formed.

Mark scheme:

• 1 mark for "completely consumed first" or "runs out first"
• 1 mark for "determines the maximum amount of product" or "stops the reaction"

Teaching note: The limiting reagent is not necessarily the reactant present in the smallest mass — it depends on the stoichiometric mole ratio. Students must compare the actual mole ratio to the required mole ratio from the balanced equation.

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Question 10 (2 marks)

Working:

• Molar mass of anhydrous $MgSO_4 = 24.3 + 32.1 + 4(16.0) = 120.4$ g mol$^{-1}$
• Mass of water in the hydrate = $246 - 120.4 = 125.6$ g mol$^{-1}$
• Molar mass of $H_2O = 2(1.0) + 16.0 = 18.0$ g mol$^{-1}$
• $x = \frac{125.6}{18.0} = 6.98 \approx 7$

Answer: $x = 7$

Teaching note: Hydrated salts contain water molecules as part of their crystal structure. The value of $x$ is found by subtracting the molar mass of the anhydrous salt from the molar mass of the hydrate, then dividing by the molar mass of water.

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Question 11

(a) (2 marks) Working:

• Molar mass of $H_2O = 2(1.0) + 16.0 = 18.0$ g mol$^{-1}$
• Mass of one molecule = $\frac{18.0}{6.02 \times 10^{23}} = 2.99 \times 10^{-23}$ g

Answer: $2.99 \times 10^{-23}$ g

Teaching note: One mole of water (18.0 g) contains $6.02 \times 10^{23}$ molecules. Dividing the molar mass by the Avogadro constant gives the mass of a single molecule.

(b) (3 marks) Working:

• Mass of 1.00 cm$^3$ of water = 1.00 g (using density)
• Moles of water: $n = \frac{1.00}{18.0} = 0.0556$ mol
• Number of molecules = $0.0556 \times 6.02 \times 10^{23} = 3.34 \times 10^{22}$

Answer: $3.34 \times 10^{22}$ molecules

Mark scheme:

• 1 mark: Use density to find mass = 1.00 g
• 1 mark: Calculate moles = 1.00/18.0
• 1 mark: Multiply by Avogadro constant for final answer

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Question 12

(a) (3 marks) Working:

• Moles of $CaCO_3$: $n = \frac{5.00}{100.1} = 0.04995$ mol (using $M_r$ of $CaCO_3 = 40.1 + 12.0 + 3(16.0) = 100.1$)
• From the equation $CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2$, mole ratio $CaCO_3 : CO_2 = 1:1$
• Moles of $CO_2 = 0.04995$ mol
• Volume of $CO_2 = 0.04995 \times 24.0 = 1.20$ dm$^3$

Answer: 1.20 dm$^3$

Mark scheme:

• 1 mark: Calculate molar mass of $CaCO_3$ and find moles
• 1 mark: Use 1:1 mole ratio
• 1 mark: Calculate volume at rtp

(b) (3 marks) Working:

• Moles of $CO_2$ collected = $\frac{0.86}{24.0} = 0.03583$ mol
• Moles of $CaCO_3$ that reacted = 0.03583 mol (1:1 ratio)
• Mass of $CaCO_3 = 0.03583 \times 100.1 = 3.587$ g
• Percentage by mass = $\frac{3.587}{5.00} \times 100 = 71.7\%$

Answer: 71.7%

Mark scheme:

• 1 mark: Moles of $CO_2$ from collected volume
• 1 mark: Mass of $CaCO_3$ calculated
• 1 mark: Percentage calculated correctly

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Question 13

(a) (2 marks) Working:

• Moles of $NH_3$: $n = \frac{34.0}{17.0} = 2.00$ mol
• From the equation: $N_2 : NH_3 = 1 : 2$, so moles of $N_2 = \frac{2.00}{2} = 1.00$ mol
• Mass of $N_2 = 1.00 \times 28.0 = 28.0$ g

Answer: 28.0 g

(b) (2 marks) Working:

• Moles of $N_2$: $n = \frac{28.0}{28.0} = 1.00$ mol
• From the equation: $N_2 : H_2 = 1 : 3$, so moles of $H_2 = 3.00$ mol
• Volume of $H_2$ at rtp = $3.00 \times 24.0 = 72.0$ dm$^3$

Answer: 72.0 dm$^3$

(c) (2 marks) Working: $\text{Percentage yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100 = \frac{17.0}{34.0} \times 100 = 50.0\%$

Answer: 50.0%

Teaching note: Percentage yield compares the actual amount of product obtained to the maximum theoretical amount. A 50% yield means half of the expected product was obtained, which could be due to incomplete reaction, side reactions, or loss during transfer.

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Question 14

(a) (2 marks) Working:

• Molar mass of NaCl = 23.0 + 35.5 = 58.5 g mol$^{-1}$
• Moles of NaCl = $\frac{29.8}{58.5} = 0.5094$ mol
• Concentration = $\frac{0.5094}{2.00} = 0.255$ mol dm$^{-3}$

Answer: 0.255 mol dm$^{-3}$

(b) (2 marks) Working: Using dilution formula: $c_1V_1 = c_2V_2$

$0.255 \times 50.0 = c_2 \times 250$

$c_2 = \frac{0.255 \times 50.0}{250} = 0.0510 \text{ mol dm}^{-3}$

Answer: 0.0510 mol dm$^{-3}$

Teaching note: When a solution is diluted, the number of moles of solute remains constant. The dilution formula $c_1V_1 = c_2V_2$ is derived from $n = cV$ where $n$ is unchanged.

(c) (3 marks) Answer:

1. Calculate the mass of NaCl needed: $n = c \times V = 0.10 \times 0.250 = 0.025$ mol; mass = $0.025 \times 58.5 = 1.46$ g
2. Weigh out 1.46 g of solid NaCl using an electronic balance
3. Dissolve the NaCl in a beaker with a small amount of distilled water, stirring with a glass rod
4. Transfer the solution to a 250 cm$^3$ volumetric flask, rinsing the beaker and glass rod with distilled water and adding the washings to the flask
5. Add distilled water until the bottom of the meniscus reaches the 250 cm$^3$ mark on the volumetric flask
6. Stopper the flask and invert several times to mix thoroughly

Mark scheme:

• 1 mark: Correct mass calculation (1.46 g)
• 1 mark: Dissolving in beaker with stirring
• 1 mark: Use of volumetric flask and making up to the mark

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Question 15

(a) (2 marks) Working:

Element	%	$A_r$	Moles	Ratio
C	85.7	12.0	7.142	1
H	14.3	1.0	14.30	2

• Moles of C = $\frac{85.7}{12.0} = 7.14$
• Moles of H = $\frac{14.3}{1.0} = 14.3$
• Ratio C : H = $7.14 : 14.3 = 1 : 2$
• Empirical formula = $CH_2$ ✓

(b) (1 mark) Working:

• Empirical formula mass of $CH_2 = 12.0 + 2(1.0) = 14.0$ g mol$^{-1}$
• Multiplier = $\frac{56.0}{14.0} = 4$
• Molecular formula = $(CH_2)_4 = C_4H_8$

Answer: $C_4H_8$

(c) (2 marks) Answer: $C_4H_8 + 6O_2 \rightarrow 4CO_2 + 4H_2O$

Mark scheme:

• 1 mark: Correct formulas for all species
• 1 mark: Correctly balanced equation

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Question 16

(a) (2 marks) Working:

• Moles of Zn = $\frac{10.0}{65.4} = 0.1529$ mol
• Moles of $H_2SO_4 = c \times V = 2.00 \times \frac{50.0}{1000} = 0.100$ mol

Answer: Zn = 0.153 mol, $H_2SO_4$ = 0.100 mol

(b) (2 marks) Answer: From the balanced equation, the mole ratio Zn : $H_2SO_4$ = 1 : 1. Since there are 0.153 mol of Zn but only 0.100 mol of $H_2SO_4$, the sulfuric acid will be consumed first. Therefore, $H_2SO_4$ is the limiting reagent.

Mark scheme:

• 1 mark: Correct identification of $H_2SO_4$ as limiting reagent
• 1 mark: Correct explanation based on mole comparison

(c) (2 marks) Working:

• From the equation: $H_2SO_4 : H_2 = 1 : 1$
• Moles of $H_2 = 0.100$ mol (limited by $H_2SO_4$)
• Volume of $H_2$ at rtp = $0.100 \times 24.0 = 2.40$ dm$^3$

Answer: 2.40 dm$^3$

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Question 17

(a) (3 marks) Working:

Element	%	$A_r$	Moles	Ratio
C	40.0	12.0	3.333	1
H	6.7	1.0	6.700	2
O	53.3	16.0	3.331	1

• Moles of C = $\frac{40.0}{12.0} = 3.33$
• Moles of H = $\frac{6.7}{1.0} = 6.7$
• Moles of O = $\frac{53.3}{16.0} = 3.33$
• Ratio C : H : O = $3.33 : 6.7 : 3.33 = 1 : 2 : 1$
• Empirical formula = $CH_2O$

Mark scheme:

• 1 mark: Correct moles calculation for each element
• 1 mark: Correct ratio simplification
• 1 mark: Correct empirical formula

(b) (1 mark) Working:

• Empirical formula mass of $CH_2O = 12.0 + 2(1.0) + 16.0 = 30.0$ g mol$^{-1}$
• Multiplier = $\frac{60.0}{30.0} = 2$
• Molecular formula = $(CH_2O)_2 = C_2H_4O_2$

Answer: $C_2H_4O_2$

(c) (2 marks) Answer: The compound is ethanoic acid, $CH_3COOH$.

$2CH_3COOH + Na_2CO_3 \rightarrow 2CH_3COONa + H_2O + CO_2$

Mark scheme:

• 1 mark: Correct identification as ethanoic acid (or $CH_3COOH$)
• 1 mark: Balanced equation with correct formulas

Teaching note: The reaction with sodium carbonate producing $CO_2$ (which turns limewater milky) is a characteristic test for carboxylic acids. The molecular formula $C_2H_4O_2$ is consistent with ethanoic acid.

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Question 18

(a) (4 marks) Working:

• Moles of $O_2 = \frac{7.20}{24.0} = 0.300$ mol
• From the equation: $2KClO_3 \rightarrow 2KCl + 3O_2$
• Mole ratio $KClO_3 : O_2 = 2 : 3$
• Moles of $KClO_3 = \frac{2}{3} \times 0.300 = 0.200$ mol
• Molar mass of $KClO_3 = 39.1 + 35.5 + 3(16.0) = 122.6$ g mol$^{-1}$
• Mass of $KClO_3 = 0.200 \times 122.6 = 24.5$ g

Answer: 24.5 g

Mark scheme:

• 1 mark: Moles of $O_2$ calculated
• 1 mark: Correct mole ratio applied (2:3)
• 1 mark: Molar mass of $KClO_3$ calculated
• 1 mark: Final mass correct

(b) (2 marks) Working: $\text{Percentage yield} = \frac{\text{actual volume}}{\text{theoretical volume}} \times 100 = \frac{6.00}{7.20} \times 100 = 83.3\%$

Answer: 83.3%

Teaching note: For gas-producing reactions, percentage yield can be calculated using volumes directly (since volume is proportional to moles at the same temperature and pressure), avoiding the need to convert to mass.

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Question 19

(a) (1 mark) Answer: $Zn(s) + 2HCl(aq) \rightarrow ZnCl_2(aq) + H_2(g)$

(b) (1 mark) Working: $n(HCl) = c \times V = 1.00 \times \frac{25.0}{1000} = 0.0250 \text{ mol}$

Answer: 0.0250 mol

(c) (2 marks) Working:

• From the equation: $HCl : H_2 = 2 : 1$
• Moles of $H_2 = \frac{0.0250}{2} = 0.0125$ mol
• Volume of $H_2$ at rtp = $0.0125 \times 24.0 = 0.30$ dm$^3$

Answer: 0.30 dm$^3$

(d) (2 marks) Answer: In Experiments 3, 4, and 5, the volume of hydrogen gas is the same (0.60 dm$^3$), which is the maximum volume that can be produced from the given amount of HCl. This indicates that HCl is the limiting reagent in these experiments — it is completely consumed, and any additional zinc does not produce more hydrogen gas. The reaction stops when all the HCl has reacted, regardless of how much excess zinc is present.

Mark scheme:

• 1 mark: HCl is the limiting reagent / is completely consumed
• 1 mark: Excess zinc does not increase the volume of $H_2$

(e) (2 marks) Answer: In Experiments 1 and 2, zinc is the limiting reagent. This can be seen because the volume of hydrogen gas increases proportionally with the mass of zinc added (0.24 dm$^3$ for 0.65 g and 0.48 dm$^3$ for 1.30 g — doubling the mass doubles the volume). This proportional relationship indicates that zinc is being completely consumed while HCl is in excess.

Mark scheme:

• 1 mark: Correct identification of Experiments 1 and 2
• 1 mark: Explanation based on proportional increase in volume with mass of Zn

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Question 20

(a) (1 mark) Working: $n(CO_2) = \frac{0.286}{24.0} = 0.01192 \text{ mol}$

Answer: 0.0119 mol

(b) (1 mark) Working: From the equation: $MCO_3 : CO_2 = 1 : 1$ $n(MCO_3) = n(CO_2) = 0.01192 \text{ mol}$

Answer: 0.0119 mol

(c) (1 mark) Working: $M(MCO_3) = \frac{m}{n} = \frac{1.50}{0.01192} = 125.8 \text{ g mol}^{-1}$

Answer: 126 g mol$^{-1}$

(d) (2 marks) Working:

• $M(M) + M(C) + 3 \times M(O) = 125.8$
• $M(M) + 12.0 + 48.0 = 125.8$
• $M(M) = 125.8 - 60.0 = 65.8$

The metal M is zinc ($A_r = 65.4$), which is the closest match.

Answer: Zinc (Zn)

Mark scheme:

• 1 mark: Correct calculation of $M(M) = 65.8$
• 1 mark: Correct identification as zinc

(e) (2 marks) Answer: If the sample contains an inert impurity, the actual mass of $MCO_3$ in the 1.50 g sample would be less than 1.50 g. This means fewer moles of $MCO_3$ would react with HCl, producing fewer moles of $CO_2$. Therefore, the volume of $CO_2$ produced would be less than that from a pure 1.50 g sample of $MCO_3$.

Mark scheme:

• 1 mark: Volume of $CO_2$ would be less
• 1 mark: Explanation — less $MCO_3$ present to react, so less $CO_2$ produced

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Total: 50 marks