Question

How many grams of phosphine (PH3) can form when 35.8 g of phosphorus and 80.3 L of hydrogen gas react at STP?

P4(s) + H2(g) → PH3(g) (Unbalanced)

_________g PH3

Answer 1

Balanced reaction is

P₄ + 6H₂   4PH₃

first calculate limiting reactant

1 mole gas occupy volume at STP = 22.414 L then 80.3 L H₂ = 80.3 / 22.414 = 3.58 mole

molar mass of P₄ = 123.9 gm/mole then 35.8 gm P₄ = 35.8 / 123.9 = 0.29 mole

According to reaction 1 mole P₄ react with 6 mole H₂ then to react with 3.58 mole H₂ required P₄ = 3.58 X 1/ 6 = 0.6 mole but P₄ given only 0.29 mole therefore P₄ is limiting reactant

According to reaction 1 mole P₄ give 4 mole PH₃ then 0,29 mole P₄ give 0,29 X 4 = 1.16 mole PH₃

molar mass of PH₃ = 33.99758 gm/mole then 1.16 mole PH₃ = 1.16 X 33.99758 = 39.437 gm

39.437 gm PH₃ produced