A Level H2 Chemistry Stoichiometry Moles Quiz

<!-- TuitionGoWhere generation metadata: stage=5-1; model=google/gemma-4-31b-it; model_label=Gemma 4 31B; generated=2026-05-27; Sources: Stage 4-0 LLM templates, syllabus context, and Stage 2 evidence where available. -->

A-Level Chemistry H2 Quiz - Stoichiometry Moles

Name: ____________________ Class: ____________________ Date: ____________________ Score: ________ / 50

Duration: 60 Minutes
Total Marks: 50
Instructions: Answer all questions. Show all working for calculations. Use the Data Booklet where necessary. Give your answers to 3 significant figures unless otherwise stated.

---

Section A: Basic Mole Calculations & Empirical Formulae (Questions 1-5)

1. Calculate the mass of $\text{Al}_2(\text{SO}_4)_3$ required to prepare $250\text{ cm}^3$ of a $0.100\text{ mol dm}^{-3}$ solution. [2]
   
   
   
   Answer: ____________________

2. A compound contains $40.0\%$ carbon, $6.7\%$ hydrogen, and $53.3\%$ oxygen by mass. Determine its empirical formula. [2]
   
   
   
   Answer: ____________________

3. A $1.50\text{ g}$ sample of a hydrated salt, $\text{MgSO}_4 \cdot x\text{H}_2\text{O}$, is heated until all water is removed. The mass of the anhydrous salt remaining is $0.60\text{ g}$. Calculate the value of $x$. [3]
   
   
   
   Answer: ____________________

4. Calculate the number of atoms of oxygen present in $0.250\text{ mol}$ of $\text{K}_2\text{Cr}_2\text{O}_7$. [2]
   
   
   
   Answer: ____________________

5. What volume of $0.500\text{ mol dm}^{-3}$ $\text{NaOH}$ is required to completely neutralize $20.0\text{ cm}^3$ of $0.200\text{ mol dm}^{-3}$ $\text{H}_2\text{SO}_4$? [2]
   
   
   
   Answer: ____________________

---

Section B: Gas Laws & Ideal Gas Equation (Questions 6-10)

6. Calculate the volume occupied by $0.120\text{ mol}$ of $\text{NH}_3$ gas at $100^\circ\text{C}$ and $101.3\text{ kPa}$. [2]
   
   
   
   Answer: ____________________

7. A gaseous hydrocarbon $\text{C}_x\text{H}_y$ has a relative molecular mass of $44.0$. At $25^\circ\text{C}$ and $1.00\text{ atm}$, $0.500\text{ g}$ of this gas occupies $276\text{ cm}^3$. Determine the molecular formula. [3]
   
   
   
   Answer: ____________________

8. $2.00\text{ g}$ of a metal $M$ reacts with excess $\text{HCl}$ to produce $400\text{ cm}^3$ of $\text{H}_2$ gas at r.t.p. ($24.0\text{ dm}^3\text{ mol}^{-1}$). Identify the metal $M$. [3]
   
   
   
   Answer: ____________________

9. Calculate the density of $\text{CO}_2$ gas at $27^\circ\text{C}$ and $1.00\text{ atm}$ in $\text{g dm}^{-3}$. [3]
   
   
   
   
   Answer: ____________________

10. A mixture of $\text{He}$ and $\text{Ne}$ has an average molar mass of $24.0\text{ g mol}^{-1}$. Calculate the mole fraction of $\text{He}$ in the mixture. [3]
    
    
    
    Answer: ____________________

---

Section C: Titrations & Limiting Reagents (Questions 11-15)

11. $25.0\text{ cm}^3$ of a solution of $\text{Na}_2\text{CO}_3$ requires $18.50\text{ cm}^3$ of $0.100\text{ mol dm}^{-3}$ $\text{HCl}$ for complete neutralization. Calculate the concentration of $\text{Na}_2\text{CO}_3$. [3]
    
    
    
    Answer: ____________________

12. In a reaction, $0.100\text{ mol}$ of $\text{AgNO}_3$ is reacted with $0.150\text{ mol}$ of $\text{NaCl}$. (a) Identify the limiting reagent. [1] (b) Calculate the mass of $\text{AgCl}$ precipitate formed. [2]
    
    
    
    Answer: ____________________

13. A $1.00\text{ g}$ sample of an impure $\text{KClO}_3$ powder is heated to decompose into $\text{KCl}$ and $\text{O}_2$. If $0.200\text{ g}$ of $\text{O}_2$ is evolved, calculate the percentage purity of the sample. [4]
    
    
    
    
    Answer: ____________________

14. $50.0\text{ cm}^3$ of $0.100\text{ mol dm}^{-3}$ $\text{Ba(OH)}_2$ is mixed with $50.0\text{ cm}^3$ of $0.100\text{ mol dm}^{-3}$ $\text{HCl}$. Calculate the final $\text{pH}$ of the solution. [4]
    
    
    
    
    Answer: ____________________

15. Calculate the mass of $\text{Fe}_2\text{O}_3$ that can be produced by reacting $10.0\text{ g}$ of $\text{Fe}$ with excess $\text{O}_2$. [3]
    
    
    
    Answer: ____________________

---

Section D: Advanced Stoichiometry & Electrolysis (Questions 16-20)

16. A current of $2.00\text{ A}$ is passed through a solution of $\text{CuSO}_4$ for $30.0$ minutes. Calculate the mass of $\text{Cu}$ deposited at the cathode. [3]
    
    
    
    Answer: ____________________

17. How many Faradays of electricity are required to reduce $0.500\text{ mol}$ of $\text{MnO}_4^-$ to $\text{Mn}^{2+}$? [2]
    
    
    
    Answer: ____________________

18. A sample of $\text{FeS}$ is analyzed. $1.00\text{ g}$ of the sample is dissolved in acid and the resulting solution is titrated against $0.100\text{ mol dm}^{-3}$ $\text{KMnO}_4$. If $20.0\text{ cm}^3$ of $\text{KMnO}_4$ is used, calculate the percentage of $\text{Fe}$ in the sample. [5]
    
    
    
    Answer: ____________________

19. Calculate the volume of $\text{H}_2$ gas (at r.t.p.) produced when $2.00\

    \
    \
    \
    **Answer:** ____________________

20. A $0.500\text{ g}$ sample of a metal oxide $\text{M}_x\text{O}_y$ is reduced by $\text{H}_2$ gas to give $0.420\text{ g}$ of metal $\text{M}$. If the relative atomic mass of $\text{M}$ is $55.8$, determine the formula of the oxide. [5]
    \
    \
    \
    **Answer:** ____________________

<!-- TuitionGoWhere generation metadata: stage=5-1; model=google/gemma-4-31b-it; model_label=Gemma 4 31B; generated=2026-05-27; Sources: Stage 4-0 LLM templates, syllabus context, and Stage 2 evidence where available. -->

Answer Key: A-Level Chemistry H2 Quiz - Stoichiometry Moles

Section A

1. $\text{mol} = 0.250 \times 0.100 = 0.0250\text{ mol}$. $\text{Mass} = 0.0250 \times 342.15 = 8.55\text{ g}$.
2. $\text{C}: 40/12 = 3.33$; $\text{H}: 6.7/1 = 6.7$; $\text{O}: 53.3/16 = 3.33$. Ratio $1:2:1 \rightarrow \text{CH}_2\text{O}$.
3. $\text{Mass water} = 1.50 - 0.60 = 0.90\text{ g}$. $\text{mol } \text{MgSO}_4 = 0.60/120.3 = 0.00499\text{ mol}$. $\text{mol } \text{H}_2\text{O} = 0.90/18.0 = 0.0500\text{ mol}$. $x = 0.0500/0.00499 \approx 10$.
4. $\text{mol O} = 0.250 \times 7 = 1.75\text{ mol}$. $\text{Atoms} = 1.75 \times 6.02 \times 10^{23} = 1.05 \times 10^{24}$.
5. $\text{mol } \text{H}_2\text{SO}_4 = 0.020 \times 0.200 = 0.00400\text{ mol}$. $\text{mol } \text{NaOH} = 2 \times 0.00400 = 0.00800\text{ mol}$. $\text{Vol} = 0.00800/0.500 = 0.0160\text{ dm}^3 = 16.0\text{ cm}^3$.

Section B

6. $V = nRT/P = (0.120 \times 8.31 \times 373) / 101.3 = 3.68\text{ dm}^3 = 3680\text{ cm}^3$.
7. $n = PV/RT = (1.00 \times 0.276) / (0.0821 \times 298) = 0.0113\text{ mol}$. $M = 0.500/0.0113 = 44.2\text{ g/mol}$. Formula: $\text{C}_3\text{H}_8$ (Propane) or $\text{C}_2\text{H}_4\text{O}$. Given hydrocarbon, $\text{C}_3\text{H}_8$ is closest but $M=44$ is $\text{C}_3\text{H}_8$ (44.1) or $\text{C}_2\text{H}_4\text{O}$ (not hydrocarbon). Correct: $\text{C}_3\text{H}_8$ (if $M=44$).
8. $n(\text{H}_2) = 0.400/24.0 = 0.0167\text{ mol}$. If $M + 2\text{HCl} \rightarrow \text{MCl}_2 + \text{H}_2$, $n(M) = 0.0167$. $Ar(M) = 2.00/0.0167 = 120$ (Tin, $\text{Sn}$).
9. $\rho = PM/RT = (1.00 \times 44.0) / (0.0821 \times 300) = 1.78\text{ g dm}^{-3}$.
10. $24 = (4 \times x) + (20.2 \times (1-x)) \rightarrow 24 = 4x + 20.2 - 20.2x \rightarrow 3.8 = -16.2x$. (Wait, He is 4, Ne is 20.2. Average 24 is impossible). Correction: If average is 15, $x=0.5$. If average is 24, the gas must be heavier than Ne. Assuming a typo in question, if $M_{avg}=12$, $x = (20.2-12)/(20.2-4) = 0.506$.