2.14 g of ammonium chloride, NH4Cl, is warmed with excess aqueous sodium hydroxide. What volume of ammonia is released at room conditions? (Ar: N 14.0, H 1.0, Cl 35.5; molar gas volume 24.0 dm3 mol−1)
- Mr of NH4Cl = 14.0 + (4 × 1.0) + 35.5 = 53.5.
- Moles of NH4Cl = 2.14 ÷ 53.5 = 0.0400 mol.
- NH4Cl + NaOH → NaCl + NH3 + H2O, a 1 : 1 ratio, so moles of NH3 = 0.0400 mol.
- Volume = 0.0400 × 24.0 = 0.960 dm3.
Answer: 0.960 dm3 of ammonia.