Calcium phosphide reacts with water: $Ca X_{3} P X_{2} + 6 H X_{2} O 3 Ca (OH) X_{2} + 2 PH X_{3}$ . A $9.10 g$ sample of pure $Ca X_{3} P X_{2}$ (atomic masses: Ca 40, P 31) is treated with excess water and the reaction proceeds with 80% yield. Taking the molar volume at STP as $22.4 L mo l^{- 1}$ , calculate the volume of phosphine obtained at STP in litres (give the answer to two decimal places).

Numerical answer type, answer in L, tolerance 0.01.

Answer: 1.79 L

Molar mass of

Ca X_{3} P X_{2}

3 (40) + 2 (31) = 182

, so moles =

9.10/182 = 0.050

. Stoichiometry gives

2 \times 0.050 = 0.100

mol

PH X_{3}

at 100% yield; at 80% this is 0.080 mol. Volume =

0.080 \times 22.4 = 1.79 L