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

= 0.28 L of PH₃

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

h let me think

the answer is the first option

explanation:

just googled it m8. not that hard

lol