56.0 L of nitrogen gas is mixed with excess hydrogen gas and it is found that 20 L of ammonia gas is produced. The volume of unused nitrogen gas is found to be _________ L.

Solution

Chemistry ma’am · 3 steps

Gas volumes react in the same ratio as moles, so 20 L of ammonia needed only 10 L of nitrogen.

1. Balanced equation. N₂ + 3H₂ → 2NH₃. At the same temperature and pressure, gas volumes are in the same ratio as moles (Gay-Lussac's law).

   V_(N₂) used = 1/2 × V_(NH₃) = 1/2 × 20 L = 10 L

2. Subtract from start. Unused N₂ = 56 - 10 = 46 L. Hydrogen is in excess, so it does not limit anything here.
3. So. The volume of unused nitrogen gas is 46 L.

The common slip: 16. Reading 1 : 2 the wrong way round says 40 L of N₂ was used, and that gives 16 L left over, not 46.

The idea to keep. For gases at the same T and P, the coefficients of the balanced equation are volume ratios. Use the product formed to find how much reactant was used, then subtract.

Answer from the NTA final answer key. Solution by PrepWiser.