What volume of phosphine (PH3) is formed by the interaction of 200 grams of calcium phosphide with hydrochloric acid?

Let’s solve the problem, make up the equation:
Ca3P2 + 6HCl = 2PH3 + 3CaCl2 – an exchange reaction during which calcium phosphide interacts with hydrochloric acid, phosphine is released;
M (Ca3P2) = 180 g / mol;
Determine the amount of moles of calcium phosphide if the mass is known:
Y (Ca3P2) = m / M = 200/180 = 1.11 mol.
Let’s make the proportion:
1.11 mol (Ca3P2) – X mol (PH3);
-1 mol                    -2 mol hence, X mol (PH3) = 1.11 * 2/1 = 2.22 mol.
Find the amount of phosphine:
V (PH3) = 2.22 * 22.4 = 49.72 HP
Answer: the volume of phosphine is 49.72 liters.