Question

# How many grams of phosphine (PH3) can form when 44.4 g of phosphorus and 98.8 L...

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)

The most common allotrope of phosphorus is P4 (tetrahedron structure)

P4 + 6 H2 --> 4 PH3

Moles of P4 = 44.4 g / 123.88 g/mol = 0.3584 mol

Moles of H2 = 98.8 L / 22.4 L/mol = 4.41 mol

Assumed new definition of STP as 1 bar (100 kPa), vice old definition at 1 atm (101.325 kPa).

We now must determine which reagent is limiting and which is in excess.

For all phosphorus to react, we need 6 times as many moles of hydrogen gas.
6 * 0.224 mol = 1.344 mol
We have more than enough hydrogen.

Phosphorus is limiting and hydrogen is in excess. We base the calculation on the phosphorus.

Moles of PH3 produced = 4 * 0.224 mol = 0.896 mol

FInally, determine mass of PH3:

m = 0.896 mol * 34.00 g/mol = 26.3 g

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