Question

How many grams of phosphine (PH3) can form when 32.0 g of phosphorus and 88.7 L of hydrogen gas react at STP? P4(s) + H2(g) → PH3(g) (Unbalanced)

g PH3?

Answer 1

P₄(s) + 6H₂(g) → 4PH₃(g)

At STP one mole of any gas occupies 22.4 L
moles H₂ = 88.7 L / 22.4 =3.96

molar mass P₄ = 123.895 g/mol
Moles P₄ = 32 g / 123.895 g/mol= 0.258

the ratio between P₄ and H₂ is 1 : 6 so moles H2 required = 0.258 x 6 =1.548

we have 3.96 moles of H₂ so H₂ is in excess and P4 is the limiting reactant

The ratio between P₄ ( limiting reactant) and PH₃ is 1: 4

moles PH₃ = 4 x 0.258 =1.032

mass PH₃ = 1.032 mol x 33.9978 g/mol=35.1 g