Secondary 3 Chemistry Stoichiometry Moles Quiz

Secondary 3 Chemistry Quiz - Stoichiometry Moles (Answer Key)

Total Marks: 40

Section A: Multiple Choice & Short Concepts

1. B [1]

• Reasoning: The mole is defined based on the number of atoms in 12g of Carbon-12.

2. B [1]

• Reasoning: 1 mole of $O_2$ contains $2 \times L$ atoms. 0.5 moles contains $0.5 \times 2 \times L = 1.0 L$ atoms.

3. A [1]

• Reasoning: Moles = Mass / $M_r$.
  • $H_2$: $1/2 = 0.5$ mol
  • $He$: $1/4 = 0.25$ mol
  • $CH_4$: $1/16 = 0.0625$ mol
  • $O_2$: $1/32 = 0.03125$ mol
  • Highest moles = Highest number of molecules.

4. C [1]

• Reasoning: $(14 + 4\times1)\times2 + 32 + 4\times16 = 36 + 32 + 64 = 132$.

5. C [1]

• Reasoning: Empirical mass of $CH_2O = 12+2+16 = 30$. Ratio $180/30 = 6$. Formula is $C_6H_{12}O_6$.

6. The reactant that is completely used up first in a chemical reaction, limiting the amount of product formed. [1]

7. [2]

• Moles of $NaOH = 4.0 / (23+16+1) = 4.0 / 40 = 0.1$ mol.
• Volume in dm³ $= 250 / 1000 = 0.25$ dm³.
• Concentration $= 0.1 / 0.25 = 0.4$ mol/dm³.
• Answer: 0.4 mol/dm³

8. Chlorine exists as a mixture of isotopes (mainly Cl-35 and Cl-37). The relative atomic mass is the weighted average of these isotopes based on their abundance. [1]

9. C [1]

• Reasoning: $M_r(Fe_2O_3) = 160$. Moles $= 160/160 = 1$ mol. Requires 1.5 mol $O_2$? No, equation is $2Mg + O_2$. Wait, question is Mg + O2.
  • Equation: $2Mg + O_2 \rightarrow 2MgO$.
  • Moles Mg $= 48/24 = 2$ mol.
  • Moles $O_2 = 32/32 = 1$ mol.
  • Ratio required Mg:O2 is 2:1. We have 2:1. It is stoichiometric.

10. [1]

• Volume $= \text{moles} \times 24 = 0.2 \times 24 = 4.8$ dm³.
• Answer: 4.8 dm³

Section B: Structured Calculations

11. (a) [3]

• $M_r(Fe_2O_3) = (56\times2) + (16\times3) = 112 + 48 = 160$.
• Moles $Fe_2O_3 = 160 \text{ g} / 160 \text{ g/mol} = 1.0$ mol.
• From equation, 1 mol $Fe_2O_3$ produces 2 mol $Fe$.
• Moles $Fe = 2.0$ mol.
• Mass $Fe = 2.0 \times 56 = 112$ g.
• Answer: 112 g

(b) [2]

• From equation, 1 mol $Fe_2O_3$ reacts with 3 mol $CO$.
• Moles $CO = 3.0$ mol.
• Volume $CO = 3.0 \times 24 = 72$ dm³.
• Answer: 72 dm³

12. (a) [1]

• Mass hydrated salt $= 25.50 - 21.50 = 4.00$ g.
• Mass anhydrous salt $= 23.90 - 21.50 = 2.40$ g.
• Mass water $= 4.00 - 2.40 = 1.60$ g.
• Answer: 1.60 g

(b) [2]

• $M_r(CuSO_4) = 64 + 32 + (16\times4) = 160$.
• Moles $CuSO_4 = 2.40 / 160 = 0.015$ mol.
• Answer: 0.015 mol

(c) [2]

• Moles $H_2O = 1.60 / 18 = 0.0888...$ mol.
• Ratio $H_2O : CuSO_4 = 0.0888... / 0.015 \approx 5.92$.
• Rounding to nearest whole number, $x = 6$.
• Answer: x = 6

13. (a) [1]

• Moles $Zn = 6.5 / 65 = 0.1$ mol.
• Answer: 0.1 mol

(b) [1]

• Volume $= 100 \text{ cm}^3 = 0.1 \text{ dm}^3$.
• Moles $HCl = 2.0 \times 0.1 = 0.2$ mol.
• Answer: 0.2 mol

(c) [2]

• Equation ratio Zn : HCl is 1 : 2.
• We have 0.1 mol Zn, which requires $0.1 \times 2 = 0.2$ mol HCl.
• We have exactly 0.2 mol HCl.
• Correction/Note: In many exam contexts, if amounts are exactly stoichiometric, neither is in excess. However, if forced to choose or if slight impurity is assumed, usually one is limiting. Here, they are stoichiometric.
• Alternative Interpretation for Exam Logic: If the question implies one must be limiting/excess, check calculations. $0.1$ mol Zn needs $0.2$ mol HCl. We have $0.2$ mol HCl. They are equivalent.
• Let's adjust the question logic for a clear answer: If the acid was 1.5 mol/dm³, moles HCl = 0.15. Then HCl is limiting. With 2.0 mol/dm³, it is exact.
• Standard Answer for Exact Stoichiometry: "Neither is in excess; they are in stoichiometric proportions." OR if the question implies a practical scenario where Zn is solid and acid is liquid, often the solid is considered the limiting factor for reaction completion if surface area is an issue, but chemically they are equal.
• Let's assume the question expects identification of limiting reactant if amounts were different. Given the numbers, the answer is: Neither / Stoichiometric amounts.
• However, to fit the "Limiting Reactant" template pattern: Let's assume the student might calculate based on a slight variation. If forced, Zn is the solid added to the solution.
• Refined Answer: Moles Zn = 0.1. Moles HCl = 0.2. Ratio 1:2. They react completely. Limiting Reactant: Neither (or both limit each other).
• Note for Marker: If student says Zn or HCl, check working. If working shows 1:2 ratio match, award marks for explanation.

(d) [2]

• Moles $H_2$ produced = Moles Zn reacted = 0.1 mol.
• Volume $H_2 = 0.1 \times 24 = 2.4$ dm³.
• Answer: 2.4 dm³

14. (a) [3]

• Assume 100 g sample.
• Mass C = 85.7 g. Moles C $= 85.7 / 12 = 7.14$ mol.
• Mass H = 14.3 g. Moles H $= 14.3 / 1 = 14.3$ mol.
• Ratio C : H $= 7.14 : 14.3$.
• Divide by smallest: $1 : 2$.
• Empirical Formula: $CH_2$.

(b) [1]

• Empirical mass $CH_2 = 14$.
• $M_r = 56$.
• Factor $= 56 / 14 = 4$.
• Molecular Formula: $C_4H_8$.

Section C: Application & Analysis

15. (a) [1]

• $M_r(MgCO_3) = 24 + 12 + 48 = 84$.
• Moles $= 2.0 / 84 = 0.0238$ mol.
• Answer: 0.0238 mol (or 0.024)

(b) [1]

• Volume $= 0.05$ dm³.
• Moles $HNO_3 = 1.0 \times 0.05 = 0.05$ mol.
• Answer: 0.05 mol

(c) [2]

• Equation ratio $MgCO_3 : HNO_3$ is 1 : 2.
• Moles $HNO_3$ needed for 0.0238 mol $MgCO_3 = 0.0238 \times 2 = 0.0476$ mol.
• We have 0.05 mol $HNO_3$.
• $0.05 > 0.0476$, so $HNO_3$ is in excess.
• Answer: Nitric Acid ($HNO_3$)

(d) [3]

• Limiting reactant is $MgCO_3$ (0.0238 mol).
• Ratio $MgCO_3 : CO_2$ is 1 : 1.
• Moles $CO_2$ expected $= 0.0238$ mol.
• Theoretical Mass $CO_2 = 0.0238 \times 44 = 1.047$ g.
• Percentage Yield $= (\text{Actual} / \text{Theoretical}) \times 100$.
• Yield $= (0.48 / 1.047) \times 100 = 45.8\%$.
• Answer: 45.8% (Accept 45-46%)

16. [1]

• Carbon dioxide ($CO_2$) is a gas and escapes into the atmosphere from the open beaker, causing a loss in mass.

17. [1]

• Any one of:
  • Incomplete reaction.
  • Loss of product during transfer/filtration.
  • Side reactions occurring.
  • Impure reactants.

18. [2]

• $M_r(H_2SO_4) = 2 + 32 + 64 = 98$.
• Concentration (g/dm³) $= \text{Molarity} \times M_r$.
• $0.5 \times 98 = 49$ g/dm³.
• Answer: 49 g/dm³

19. [3]

• Moles $NaOH = (20/1000) \times 0.1 = 0.002$ mol.
• Ratio $NaOH : H_2SO_4$ is 2 : 1.
• Moles $H_2SO_4 = 0.002 / 2 = 0.001$ mol.
• Volume $H_2SO_4 = 25 \text{ cm}^3 = 0.025 \text{ dm}^3$.
• Concentration $H_2SO_4 = 0.001 / 0.025 = 0.04$ mol/dm³.
• Answer: 0.04 mol/dm³

20. [1]

• Percentage purity is the percentage by mass of the pure compound in a sample containing impurities.