Question

What is the maximum number of grams of PH₃ that can be formed when 6.2 g of phosphorus reacts with 4.0 g of hydrogen to form PH₃?
`P_4+6H_2 \rightarrow 4PH_3`

Answer 1

The maximum number of grams of PH₃ that can be formed is 6.79g

Step-by-step explanation:

This is the reaction:

P₄ + 6H₂ → 4PH₃

Let's find out the limiting reactant so we can work with.

1 mol of P₄ = 123.88 g/m

6.2 g / 123.88 g/m = 0.05 moles

1 mof of H₂ = 2 g/m

4 g / 2g/m = 2 mol

Clearly the limiting reactant is the P₄ but let's demonstrate it

6 mol of hydrogen react with 1 mol of phosphorus

2 mol of hydrogen react with (2 .1)/6 = 0.333 moles of P₄

We don't have enough P₄, that's why it is the limiting.

1 mol of P₄ produce 4 moles of phosphine

0.05 moles of P₄ will produce, the quadruple = 0.2 moles

Let's determine the grams (moles . molar mass)

0.2 moles . 33.97 g/m = 6.79 g