Questions: Gas Stoichiometry and Volume-Volume Calculations

Question 1

Consider the reaction: 2H₂(g) + O₂(g) → 2H₂O(g). All gases are at the same temperature and pressure. If you have 6.0 L of H₂, what volume of O₂ is needed for complete reaction?

• A. 6.0 L, because gas volumes always react in 1:1 ratios at the same conditions
• B. 3.0 L, because the volume ratio equals the mole ratio from the balanced equation
• C. 12.0 L, because oxygen is heavier and you need more volume to provide the same mass
• D. You must convert to moles first — volume ratios cannot be used directly

Answer: (B)

When all gases are at the same temperature and pressure, volume ratios equal mole ratios (Avogadro's law). The balanced equation shows a 2:1 mole ratio of H₂ to O₂, so the volume ratio is also 2:1. Six liters of H₂ requires 3.0 L of O₂. No mole conversion is needed. The mass or molar mass of the gases is irrelevant; at the same T and P, equal volumes always contain equal moles — that is Avogadro's law.

Question 2

Zinc metal reacts with hydrochloric acid: Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g). You have 6.54 g of zinc at STP. What is the correct approach to find the volume of H₂ produced?

• A. Use the 1:1 mole ratio directly as a volume ratio to convert grams of Zn to liters of H₂
• B. Convert Zn grams to moles using molar mass, apply the mole ratio, then use 22.4 L/mol at STP to find liters of H₂
• C. Use PV = nRT with the mass of zinc to find the volume directly
• D. Volume ratios cannot be used here at all — molarity of HCl is needed instead

Answer: (B)

When a solid (or liquid) is involved, the gas volume shortcut does not apply — there is no volume to ratio for solid zinc. You must follow the full stoichiometric chain: (1) convert grams of Zn to moles using molar mass (65.4 g/mol → 0.100 mol Zn), (2) apply the 1:1 mole ratio to get 0.100 mol H₂, (3) convert to volume at STP using 22.4 L/mol → 2.24 L H₂. The volume-ratio shortcut only works when all participants in the ratio are gases at the same T and P.

Question 3

For the reaction N₂(g) + 3H₂(g) → 2NH₃(g), all gases at the same temperature and pressure: 10 L of N₂ reacts completely with 30 L of H₂ to produce 20 L of NH₃.

• T. True
• F. False

Answer: True

At the same T and P, volume ratios equal mole ratios. The balanced equation shows a 1:3:2 mole ratio, so the volume ratio is also 1:3:2. Ten liters of N₂ requires 30 L of H₂ and produces 20 L of NH₃. Notice that the total gas volume decreases from 40 L of reactants to 20 L of product — gas-phase reactions can and do change the total volume of gas in the system.

Question 4

If 5 L of gas A is mixed with 5 L of gas B at the same temperature and pressure and they react mostly with each other, then the balanced equation should have a 1:1 mole ratio of A to B.

• T. True
• F. False

Answer: False

The volume ratio reflects the mole ratio only if both volumes represent the amounts that actually reacted. If 5 L of A is mixed with 5 L of B and one is in excess, the volumes mixed are not equal to the volumes that reacted. You need to identify the limiting reagent from the balanced equation first. The shortcut applies when computing required volumes from the equation's coefficients, not when inferring the equation from two arbitrary mixed volumes.

Question 5

Explain when you can use the volume-ratio shortcut in gas stoichiometry and when you must convert through moles. What physical principle justifies the shortcut?

Model answer: The volume-ratio shortcut applies only when all reactants and products being compared are gases at the same temperature and pressure. In that case, volume ratios equal mole ratios, so the balanced equation's coefficients directly give volume ratios. When solids or liquids are involved (or gases at different T and P), you must convert non-gas quantities to moles using molar mass, apply the mole ratio, and then convert gas moles to volume using 22.4 L/mol at STP or PV = nRT at other conditions. The justifying principle is Avogadro's law: equal volumes of any gas at the same T and P contain equal numbers of moles.

The shortcut is a special case of the general stoichiometric approach. The general path is always: (mass or volume) → moles → moles → (mass or volume). The shortcut skips the middle steps for gas-to-gas conversions because the volume-to-moles conversion factor (RT/P) is the same for all gases and cancels when T and P are equal. Recognizing which path applies is the core skill: are all species involved gases, and are they all at the same conditions?