Gas Calculations

Amount of Substance Worksheet

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Easy Gas Calculation Questions

1. 40.0g of magnesium reacted with excess hydrochloric acid. Given that the temperature of the gas was 25^oC and the pressure was 102kPa, what volume of hydrogen gas did it produce?

Working Out

Mg + 2HCl → MgCl₂ + H₂

Step 1: Moles of Magnesium n(Mg) = 40.0 / 24.3 = 1.646 mol

Step 2: Moles of Hydrogen Ratio Mg:H₂ is 1:1
n(H₂) = 1.646 mol

Step 3: Volume Calculation (PV=nRT) P = 102,000 Pa, T = 298 K, R = 8.31
V = (n × R × T) / P
V = (1.646 × 8.31 × 298) / 102000 = 0.03997 m³
Answer = 40.0 dm³

2. What volume of bromine gas would you required to produce 150g of 1,2-dibromohexane, given the temperature of 20^oC and a pressure of 100kPa?

Working Out

C₆H₁₂ + Br₂ → C₆H₁₂Br₂

Step 1: Moles of Product Mr(C₆H₁₂Br₂) = 72.0 + 12.0 + 159.8 = 243.8
n = 150 / 243.8 = 0.6153 mol

Step 2: Volume of Bromine Ratio is 1:1, so n(Br₂) = 0.6153 mol
V = (0.6153 × 8.31 × 293) / 100000 = 0.01498 m³
Answer = 15.0 dm³

3. What volume of oxygen would you need to react with 100kg of octane, assuming that the reaction is complete combustion? The oxygen is stored at 20^oC at a pressure of 20 000kPa.

Working Out

C₈H₁₈ + 12.5O₂ → 8CO₂ + 9H₂O

Step 1: Moles of Octane Mr(C₈H₁₈) = 114.2
n = 100,000 / 114.2 = 875.66 mol

Step 2: Moles of Oxygen n(O₂) = 875.66 × 12.5 = 10945.7 mol

Step 3: Volume P = 20,000,000 Pa
V = (10945.7 × 8.31 × 293) / 20000000 = 1.33 m³
Answer = 1330 dm³ (or 1.33 m³)

4. What volume of carbon dioxide would you produce if you react 500g of calcium carbonate with excess sulfuric acid? The gas produced is at standard pressure and a temperature of 0^oC.

Working Out

CaCO₃ + H₂SO₄ → CaSO₄ + H₂O + CO₂

Step 1: Moles of CaCO₃ Mr = 100.1
n = 500 / 100.1 = 4.995 mol

Step 2: Volume P = 100,000 Pa (Standard Pressure), T = 273 K
V = (4.995 × 8.31 × 273) / 100000 = 0.1133 m³
Answer = 113 dm³

5. What volume of oxygen do you need to produce 500kg of sulfur trioxide from oxygen and sulfur when at 1 atmosphere and 25^oC?

Working Out

2S + 3O₂ → 2SO₃

Step 1: Moles of SO₃ Mr(SO₃) = 32.1 + 48.0 = 80.1
n = 500,000 / 80.1 = 6242.2 mol

Step 2: Moles of Oxygen Ratio 3:2 = 1.5
n(O₂) = 6242.2 × 1.5 = 9363.3 mol

Step 3: Volume P = 101,325 Pa (1 atm), T = 298 K
V = (9363.3 × 8.31 × 298) / 101325 = 228.8 m³
Answer = 229 m³

Harder Gas Calculation Questions

1. What volume of gas will you produce by the decomposition of 500g of hydrazine at standard pressure and 0^oC?

Working Out

N₂H₄ → N₂ + 2H₂

Assuming complete decomposition into constituent gases.

Step 1: Moles of Hydrazine Mr = 32.0
n = 500 / 32.0 = 15.625 mol

Step 2: Moles of Gas Produced 1 mol solid → 3 mol gas
n(Gas) = 15.625 × 3 = 46.875 mol

Step 3: Volume V = (46.875 × 8.31 × 273) / 100000 = 1.063 m³
Answer = 1.06 m³

2. 70.0g of barium and 500cm³ of 2moldm^-3 of hydrochloric acid react. What volume of hydrogen gas do they make at one atmosphere and 298k?

Working Out

Ba + 2HCl → BaCl₂ + H₂

Step 1: Calculate Moles n(Ba) = 70.0 / 137.3 = 0.510 mol
n(HCl) = 2 × 0.500 = 1.00 mol

Step 2: Determine Limiting Reagent Ba requires 2x mols of HCl (1.02 mol). We only have 1.00 mol.
HCl is limiting.

Step 3: Volume of Hydrogen n(H₂) = 1.00 / 2 = 0.500 mol
V = (0.500 × 8.31 × 298) / 101325 = 0.0122 m³
Answer = 12.2 dm³

3. 100 litres of liquified propane gas are reacted with 100 litres of oxygen gas. The oxygen is stored at 20 000kPa and is at 25^oC and the LPG is a liquid with a density of 0.5kg/l. What volume of carbon dioxide will you make, assuming that the reaction is complete combustion and that the carbon dioxide cools to 25^oC and one atmosphere?

Working Out

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Step 1: Moles of Propane Mass = 100 L × 0.5 kg/L = 50 kg = 50,000 g
n(Propane) = 50,000 / 44.0 = 1136 mol

Step 2: Moles of Oxygen n = PV/RT = (20,000,000 × 0.1) / (8.31 × 298) = 807.6 mol

Step 3: Limiting Reagent Need 5x oxygen for propane (5680 mol). We have 807.6.
Oxygen is severely limiting.

Step 4: Volume of CO₂ n(CO₂) = 807.6 × (3/5) = 484.6 mol
V = (484.6 × 8.31 × 298) / 101325 = 11.84 m³
Answer = 11.8 m³

4. What volume of H₂ at 20^oC and 100kPa would you require to fully react with 100cm³ of 0.5moldm^-3 tin (IV) chloride?

Working Out

SnCl₄ + 2H₂ → Sn + 4HCl

Step 1: Moles of SnCl₄ n = 0.5 × 0.100 = 0.05 mol

Step 2: Volume of H₂ n(H₂) = 0.05 × 2 = 0.10 mol
V = (0.10 × 8.31 × 293) / 100000 = 0.00243 m³
Answer = 2.43 dm³

5. 60.00dm³ of methane reacted with a limited supply of oxygen at 1 atmosphere and 25^oC, producing 29.33dm³of carbon dioxide, 9.78 dm³ of carbon monoxide, and 3.600g of carbon.

a) What volume of oxygen did the methane react with?

Reaction Stoichiometries (Volumetric) 1) CH₄ + 2O₂ → CO₂ + 2H₂O (Ratio 1:2)
2) CH₄ + 1.5O₂ → CO + 2H₂O (Ratio 1:1.5)
3) CH₄ + O₂ → C + 2H₂O (Ratio 1:1)

Solid Carbon Equivalent Volume Molar Vol at 25C = (8.31 × 298) / 101325 = 0.02444 m³ = 24.44 dm³
n(C) = 3.600 / 12.0 = 0.300 mol
Equiv Vol = 0.300 × 24.44 = 7.33 dm³

Total Oxygen For CO₂: 29.33 × 2 = 58.66 dm³
For CO: 9.78 × 1.5 = 14.67 dm³
For C: 7.33 × 1 = 7.33 dm³
Total = 58.66 + 14.67 + 7.33
Answer = 80.66 dm³

b) What volume of methane would there be left over?

Methane Used Total Carbon-containing products volume = 29.33 + 9.78 + 7.33 = 46.44 dm³
(Since 1 mol CH₄ produces 1 mol C-product, volumes equate).

Remaining Methane Remaining = Initial – Used
Remaining = 60.00 – 46.44
Answer = 13.56 dm³