Question

How many grams of phosphine (PH₃) can form when 44.4 g of phosphorus and 98.8 L of hydrogen gas react at STP?

P₄(s) + H₂(g) → PH₃(g) (Unbalanced)

Answer 1

The balanced chemical equation is: P₄(s) + 6H₂(g) 4PH₃(g)

we will solve these types of question in 3 steps:

Step-1: Converting the given amounts in terms of moles

moles of P₄ = mass of P₄/ Molar mass of P₄ = 44.4/123.88 = 0.3584 mol

moles of H₂ = mass of H₂/ Molar mass of H₂ = 98.8/22.4 = 4.4107 mol (For gases 1mol= 22.4L at STP )

Step-2: Identifying the Limiting Reagent

As from the above given balanced equation, we can see that 6 mol of H₂ reacts with 1 mol of P₄

therefore, 4.4107 mol of H2 will react P₄ = (4.4107 X 1)/6 = 0.7351 mol

But we have only 0.3584 mol of P_4. Hence, H₂ is not the limiting reagent.

Now try in the reverse direction,

As from the above given balanced equation, we can see that 1 mol of P₄ reacts with 6 mol of H₂

therefore,  0.3584 mol of P₄ will react H₂ = 0.3584 X 6 = 2.1504 mol

As we have 4.4107 mol of H_2, therefore H₂ is the excess reagent whereas P₄ is the limiting reagent.

Step-3: Calculating the amount of PH₃ formed

As the amount of product formed depends on limiting reagent, therefore

1mol of P₄ forms PH₃ = 4 mol

0.3584 mol of P₄ will form PH₃ = 0.3584 X 4 = 1.4336 mol

Now, using the formula; Moles = Mass/ Molar mass

1.4336 = Mass of PH₃ formed / 33.99

  Mass of PH₃ formed = 1.4336 X 33.99 = 48.7280 g