How many liters of phosphine are formed when 0.42 L of hydrogen react with phosphorus

Answer;

= 0.28 L of PH₃

Explanation and solution;

Phosphorus and hydrogen can be combined to form phosphine

P₄ (s) + 6H₂ (g) → 4PH₃ (g)

From the equation; we want to determine how many L of phosphine will be formed from 0.42 L of hydrogen;

1 mole of a gas occupies 22.4 L at STP

Therefore; moles of H2 = 0.42 L/22.4 L

The mole ratio of H2 to PH3 is 4 moles : 6 moles

Thus;

Moles of PH3 = 0.42/22.4 * 4/6

Hence;

Volume of PH3 = 0.42/22.4 * 4/6 * 22.4/1

= 0.28 L

Thus; we form 0.28 L of phosphine