Excess $NaOH$ solution is boiled with $6.60 g$ of ammonium sulphate, $(NH X_{4}) X_{2} SO X_{4}$ (atomic masses: H 1, N 14, O 16, S 32). Only 75% of the ammonia formed is collected. Taking the molar volume at STP as $22.4 L mo l^{- 1}$ , calculate the volume of ammonia collected at STP in litres (give the answer to two decimal places).

Numerical answer type, answer in L, tolerance 0.01.

Answer: 1.68 L

(NH X_{4}) X_{2} SO X_{4} + 2 NaOH Na X_{2} SO X_{4} + 2 NH X_{3} + 2 H X_{2} O

. Molar mass = 132, so moles =

6.60/132 = 0.050

, giving 0.100 mol

NH X_{3}

. Collected =

0.100 \times 0.75 = 0.0750

mol, volume =

0.0750 \times 22.4 = 1.68 L