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A Level H2 Chemistry Stoichiometry Moles Quiz

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Questions

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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 & Titrations

Questions 1–5 focus on molarity, titration data, and stoichiometry.

  1. Calculate the number of moles of NaOH\text{NaOH} present in 25.0 cm325.0\text{ cm}^3 of a 0.150 mol dm−30.150\text{ mol dm}^{-3} solution. [2]
    \
  2. A 20.0 cm320.0\text{ cm}^3 sample of 0.100 mol dm−3H2SO40.100\text{ mol dm}^{-3} \text{H}_2\text{SO}_4 is neutralized by 25.0 cm325.0\text{ cm}^3 of NaOH\text{NaOH}. Calculate the concentration of the NaOH\text{NaOH} solution. [3]
    \
  3. 25.0 cm325.0\text{ cm}^3 of 0.0200 mol dm−3KMnO40.0200\text{ mol dm}^{-3} \text{KMnO}_4 is used to titrate a solution of Fe2+\text{Fe}^{2+}. Calculate the mass of Fe\text{Fe} present in the sample. [4]
    \
  4. What mass of Na2CO3\text{Na}_2\text{CO}_3 is required to react completely with 25.0 cm325.0\text{ cm}^3 of 0.100 mol dm−3HCl0.100\text{ mol dm}^{-3} \text{HCl}? [3]
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  5. Calculate the mass of CH3COOH\text{CH}_3\text{COOH} that can be neutralized by 20.0 cm320.0\text{ cm}^3 of 0.100 mol dm−3NaOH0.100\text{ mol dm}^{-3} \text{NaOH}. [3]
    \

Section B: Gas Laws & Stoichiometry

Questions 6–10 focus on ideal gas laws and volumetric calculations.

  1. Calculate the mass of Na2C2O4\text{Na}_2\text{C}_2\text{O}_4 required to react with 20.0 cm320.0\text{ cm}^3 of 0.0200 mol dm−3KMnO40.0200\text{ mol dm}^{-3} \text{KMnO}_4. [4]
    \
  2. Determine the mass of NaOH\text{NaOH} that reacts completely with 25.0 cm325.0\text{ cm}^3 of 0.100 mol dm−3HCl0.100\text{ mol dm}^{-3} \text{HCl}. [3]
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  3. Calculate the number of moles of an ideal gas occupying 2.00 L2.00\text{ L} at 1 atm1\text{ atm} and 298 K298\text{ K}. [3]
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  4. What volume of H2\text{H}_2 gas (at 24.0 dm3 mol−124.0\text{ dm}^3\text{ mol}^{-1}) is produced when 0.243 g0.243\text{ g} of Mg\text{Mg} reacts with excess HCl\text{HCl}? [4]
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  5. Calculate the volume of CO2\text{CO}_2 gas (at 24.0 dm3 mol−124.0\text{ dm}^3\text{ mol}^{-1}) produced from the decomposition of 1.00 g1.00\text{ g} of CaCO3\text{CaCO}_3. [4]
    \

Section C: Empirical and Molecular Formulae

Questions 11–15 focus on compound analysis and formula determination.

  1. 0.100 mol0.100\text{ mol} of H2\text{H}_2 reacts with 0.050 mol0.050\text{ mol} of O2\text{O}_2 to form H2O\text{H}_2\text{O}. Calculate the mass of water produced. [4]
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  2. 2.70 g2.70\text{ g} of Al\text{Al} reacts with 0.200 mol0.200\text{ mol} of Cl2\text{Cl}_2. Calculate the mass of AlCl3\text{AlCl}_3 formed. [5]
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  3. A compound contains 40.0 g40.0\text{ g} of C\text{C}, 6.7 g6.7\text{ g} of H\text{H}, and 32.0 g32.0\text{ g} of O\text{O}. Determine its empirical formula. [5]
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  4. A compound has an empirical formula of CH2O\text{CH}_2\text{O} and a molar mass of 180 g mol−1180\text{ g mol}^{-1}. Determine its molecular formula. [3]
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  5. An oxide contains 55.8 g55.8\text{ g} of Fe\text{Fe} and 48.0 g48.0\text{ g} of O\text{O}. Determine the empirical formula. [4]
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Section D: Advanced Stoichiometry & Yield

Questions 16–20 focus on limiting reagents and percentage yield.

  1. Calculate the mass of O2\text{O}_2 required for the complete combustion of 44.1 g44.1\text{ g} of propane (C3H8\text{C}_3\text{H}_8). [4]
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  2. If the combustion of 44.1 g44.1\text{ g} of propane produces 110 g110\text{ g} of CO2\text{CO}_2, calculate the percentage yield. [4]
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  3. Calculate the mass of AgCl\text{AgCl} formed when 0.100 mol0.100\text{ mol} of AgNO3\text{AgNO}_3 reacts with 0.150 mol0.150\text{ mol} of NaCl\text{NaCl}. [3]
    \
  4. A 5.00 g5.00\text{ g} sample of a metal carbonate M2CO3\text{M}_2\text{CO}_3 releases 1.20 dm31.20\text{ dm}^3 of CO2\text{CO}_2 at r.t.p. Calculate the molar mass of the metal M\text{M}. [4]
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  5. 10.0 g10.0\text{ g} of CaCO3\text{CaCO}_3 is heated to produce CaO\text{CaO} and CO2\text{CO}_2. If the actual yield of CaO\text{CaO} is 4.80 g4.80\text{ g}, calculate the percentage yield. [3]
    \

Answers

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

Section A: Basic Mole Calculations & Titrations

  1. moles=0.150 mol dm−3×(25.0/1000) dm3=3.75×10−3 mol\text{moles} = 0.150\text{ mol dm}^{-3} \times (25.0 / 1000)\text{ dm}^3 = 3.75 \times 10^{-3}\text{ mol} Answer: 3.75×10−3 mol3.75 \times 10^{-3}\text{ mol} [2]

  2. moles of H2SO4=0.100×0.020=2.00×10−3 mol\text{moles of } \text{H}_2\text{SO}_4 = 0.100 \times 0.020 = 2.00 \times 10^{-3}\text{ mol} moles of NaOH=2×2.00×10−3=4.00×10−3 mol\text{moles of } \text{NaOH} = 2 \times 2.00 \times 10^{-3} = 4.00 \times 10^{-3}\text{ mol} Conc=(4.00×10−3)/0.025=0.160 mol dm−3\text{Conc} = (4.00 \times 10^{-3}) / 0.025 = 0.160\text{ mol dm}^{-3} Answer: 0.160 mol dm−30.160\text{ mol dm}^{-3} [3]

  3. moles of KMnO4=0.0200×0.025=5.00×10−4 mol\text{moles of } \text{KMnO}_4 = 0.0200 \times 0.025 = 5.00 \times 10^{-4}\text{ mol} moles of Fe2+=(5/2)×5.00×10−4=1.25×10−3 mol\text{moles of } \text{Fe}^{2+} = (5/2) \times 5.00 \times 10^{-4} = 1.25 \times 10^{-3}\text{ mol} Mass=1.25×10−3×55.8=0.0698 g\text{Mass} = 1.25 \times 10^{-3} \times 55.8 = 0.0698\text{ g} Answer: 0.0698 g0.0698\text{ g} [4]

  4. moles of HCl=0.100×0.025=2.50×10−3 mol\text{moles of } \text{HCl} = 0.100 \times 0.025 = 2.50 \times 10^{-3}\text{ mol} moles of Na2CO3=2.50×10−3/2=1.25×10−3 mol\text{moles of } \text{Na}_2\text{CO}_3 = 2.50 \times 10^{-3} / 2 = 1.25 \times 10^{-3}\text{ mol} Mass=1.25×10−3×106=0.133 g\text{Mass} = 1.25 \times 10^{-3} \times 106 = 0.133\text{ g} Answer: 0.133 g0.133\text{ g} [3]

  5. moles of NaOH=0.100×0.020=2.00×10−3 mol\text{moles of } \text{NaOH} = 0.100 \times 0.020 = 2.00 \times 10^{-3}\text{ mol} moles of CH3COOH=2.00×10−3 mol\text{moles of } \text{CH}_3\text{COOH} = 2.00 \times 10^{-3}\text{ mol} Mass=2.00×10−3×60.0=0.120 g\text{Mass} = 2.00 \times 10^{-3} \times 60.0 = 0.120\text{ g} Answer: 0.120 g0.120\text{ g} [3]

Section B: Gas Laws & Stoichiometry

  1. moles of KMnO4=0.0200×0.020=4.00×10−4 mol\text{moles of } \text{KMnO}_4 = 0.0200 \times 0.020 = 4.00 \times 10^{-4}\text{ mol} moles of Na2C2O4=(5/2)×4.00×10−4=1.00×10−3 mol\text{moles of } \text{Na}_2\text{C}_2\text{O}_4 = (5/2) \times 4.00 \times 10^{-4} = 1.00 \times 10^{-3}\text{ mol} Mass=1.00×10−3×134=0.134 g\text{Mass} = 1.00 \times 10^{-3} \times 134 = 0.134\text{ g} Answer: 0.134 g0.134\text{ g} [4]

  2. moles of HCl=0.100×0.025=2.50×10−3 mol\text{moles of } \text{HCl} = 0.100 \times 0.025 = 2.50 \times 10^{-3}\text{ mol} moles of NaOH=2.50×10−3 mol\text{moles of } \text{NaOH} = 2.50 \times 10^{-3}\text{ mol} Mass=2.50×10−3×40.0=0.100 g\text{Mass} = 2.50 \times 10^{-3} \times 40.0 = 0.100\text{ g} Answer: 0.100 g0.100\text{ g} [3]

  3. n=PV/RT=(1×2.00)/(0.0821×298)=0.0820 moln = PV / RT = (1 \times 2.00) / (0.0821 \times 298) = 0.0820\text{ mol} Answer: 0.0820 mol0.0820\text{ mol} [3]

  4. moles of Mg=0.243/24.3=0.0100 mol\text{moles of } \text{Mg} = 0.243 / 24.3 = 0.0100\text{ mol} moles of H2=0.0100 mol\text{moles of } \text{H}_2 = 0.0100\text{ mol} Volume=0.0100×24.0=0.240 dm3=240 cm3\text{Volume} = 0.0100 \times 24.0 = 0.240\text{ dm}^3 = 240\text{ cm}^3 Answer: 240 cm3240\text{ cm}^3 [4]

  5. moles of CaCO3=1.00/100=0.0100 mol\text{moles of } \text{CaCO}_3 = 1.00 / 100 = 0.0100\text{ mol} moles of CO2=0.0100 mol\text{moles of } \text{CO}_2 = 0.0100\text{ mol} Volume=0.0100×24.0=0.240 dm3=240 cm3\text{Volume} = 0.0100 \times 24.0 = 0.240\text{ dm}^3 = 240\text{ cm}^3 Answer: 240 cm3240\text{ cm}^3 [4]

Section C: Empirical and Molecular Formulae

  1. Stoichiometry: 2H2+O2→2H2O\text{Stoichiometry: } 2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O} moles of H2O=0.100 mol\text{moles of } \text{H}_2\text{O} = 0.100\text{ mol} Mass=0.100×18.0=1.80 g\text{Mass} = 0.100 \times 18.0 = 1.80\text{ g} Answer: 1.80 g1.80\text{ g} [4]

  2. moles of Al=2.70/27.0=0.100 mol\text{moles of } \text{Al} = 2.70 / 27.0 = 0.100\text{ mol} moles of Cl2=0.200 mol\text{moles of } \text{Cl}_2 = 0.200\text{ mol} Al is limiting reagent (requires 0.150 mol Cl2)\text{Al is limiting reagent (requires } 0.150\text{ mol } \text{Cl}_2) Mass of AlCl3=0.100×133.3=13.3 g\text{Mass of } \text{AlCl}_3 = 0.100 \times 133.3 = 13.3\text{ g} Answer: 13.3 g13.3\text{ g} [5]

  3. C:40/12=3.33;H:6.7/1=6.7;O:32/16=2.00\text{C}: 40/12 = 3.33; \text{H}: 6.7/1 = 6.7; \text{O}: 32/16 = 2.00 Ratio C:H:O=1.67:3.35:1≈5:10:3\text{Ratio C:H:O} = 1.67 : 3.35 : 1 \approx 5 : 10 : 3 Answer: C5H10O3\text{C}_5\text{H}_{10}\text{O}_3 [5]

  4. Empirical mass (CH2O)=30.0 g mol−1\text{Empirical mass } (\text{CH}_2\text{O}) = 30.0\text{ g mol}^{-1} n=180/30.0=6n = 180 / 30.0 = 6 Molecular formula=C6H12O6\text{Molecular formula} = \text{C}_6\text{H}_{12}\text{O}_6 Answer: C6H12O6\text{C}_6\text{H}_{12}\text{O}_6 [3]

  5. Fe:55.8/55.8=1 mol;O:48.0/16.0=3 mol\text{Fe}: 55.8/55.8 = 1\text{ mol}; \text{O}: 48.0/16.0 = 3\text{ mol} Answer: FeO3\text{FeO}_3 [4]

Section D: Advanced Stoichiometry & Yield

  1. moles of C3H8=44.1/44.1=1.00 mol\text{moles of } \text{C}_3\text{H}_8 = 44.1 / 44.1 = 1.00\text{ mol} moles of O2=5×1.00=5.00 mol\text{moles of } \text{O}_2 = 5 \times 1.00 = 5.00\text{ mol} Mass=5.00×32.0=160 g\text{Mass} = 5.00 \times 32.0 = 160\text{ g} Answer: 160 g160\text{ g} [4]

  2. Theoretical mass CO2=(3×1.00)×44.0=132 g\text{Theoretical mass } \text{CO}_2 = (3 \times 1.00) \times 44.0 = 132\text{ g} Yield=(110/132)×100=83.3%\text{Yield} = (110 / 132) \times 100 = 83.3\% Answer: 83.3%83.3\% [4]

  3. moles of AgNO3=0.100 mol\text{moles of } \text{AgNO}_3 = 0.100\text{ mol} (Limiting) moles of AgCl=0.100 mol\text{moles of } \text{AgCl} = 0.100\text{ mol} Mass=0.100×143.3=14.3 g\text{Mass} = 0.100 \times 143.3 = 14.3\text{ g} Answer: 14.3 g14.3\text{ g} [3]

  4. moles of CO2=1.20/24.0=0.050 mol\text{moles of } \text{CO}_2 = 1.20 / 24.0 = 0.050\text{ mol} moles of M2CO3=0.050 mol\text{moles of } \text{M}_2\text{CO}_3 = 0.050\text{ mol} Molar mass M2CO3=5.00/0.050=100 g mol−1\text{Molar mass } \text{M}_2\text{CO}_3 = 5.00 / 0.050 = 100\text{ g mol}^{-1} 2M+12.0+48.0=100→2M=40→M=20.0 g mol−12\text{M} + 12.0 + 48.0 = 100 \rightarrow 2\text{M} = 40 \rightarrow \text{M} = 20.0\text{ g mol}^{-1} Answer: 20.0 g mol−120.0\text{ g mol}^{-1} [4]

  5. moles of CaCO3=10.0/100=0.100 mol\text{moles of } \text{CaCO}_3 = 10.0 / 100 = 0.100\text{ mol} Theoretical mass CaO=0.100×56.0=5.60 g\text{Theoretical mass } \text{CaO} = 0.100 \times 56.0 = 5.60\text{ g} Yield=(4.80/5.60)×100=85.7%\text{Yield} = (4.80 / 5.60) \times 100 = 85.7\% Answer: 85.7%85.7\% [3]