How many moles of nitrogen would react with excess hydrogen to produce 520 ml of ammonia?

0.0116 mol.

Explanation:

It is known that at STP: 1 mol of any gas occupies 22.4 L.

Using cross multiplication:

1 mol of NH₃ occupies → 22.4 L.

??? mol of NH₃ occupies → (520 mL) 0.52 L.

∴ The no. of moles of (520 mL) NH₃ produced = (1 mol)(0.52 L)/(22.4 L) = 0.0232 mol.

3H₂ + N₂ → 2NH₃.

3 mol of H₂ react with 1 mol of N₂ to produce 2 mol of NH₃.

Using cross multiplication:

1 mol of N₂ produce → 2 mol of NH₃, from stichiometry.

??? mol of N₂ produce → 0.0232 mol of NH₃.

∴ The no. of moles of N₂ needed = (1 mol)(0.0232 mol)/(2 mol) = 0.0116 mol.

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