AlP + 3 H2O --> PH3 + Al(OH)3 If 100.0 g of aluminum phosphide is mixed with 50.0 g of water, what is the percent yield if 26.7 g of phosphine (PH3) is produced?

Question

Consider the following reaction:

AlP + 3 H₂O --> PH₃ + Al(OH)₃

If 100.0 g of aluminum phosphide is mixed with 50.0 g of water, what is the percent yield if 26.7 g of phosphine (PH3) is produced?

Answer 1

First we will want to convert the mass of AlP and H2O to moles, so we will need the MW of AlP and H2O

MW(AlP) = 58 g/mol

MW (H2O) = 18 g/mol

Then we can convert the masses to moles by dividing the mass by the MW

Mol AlP = 100.0 g / 58g/mol = 1.72 mol

Mol H2O = 50.0 g / 18g/mol = 2.78 mol

From the balanced reaction we can see that there is a 3:1 mol ratio between H2O and AlP, so we need 3x as many moles of H2O to AlP in order for all of the AlP to react. So 3 * 1.72 = 5.16, so we would need 5.16 mol of H2O in order to react all of the AlP. Since we only have 2.78 mol of H2O, which is less than 5.16, we can see that H2O is the limiting reactant in this scenario.

So we will use the mol of H2O to calculate the theoretical yield of PH3.

We can see from the balanced reaction that 1 mol of PH3 is formed for every 3 mol of H2O reacted. So the theoretical yield of PH3 formed will be 2.78 mol / 3 = 0.93 mol PH3.

We can convert this mol to mass by multiplying by the MW of PH3 = 34 g/mol

0.93 mol PH3 * 34 g/mol = 31.5 grams of PH3 (theoretical )

Percent Yield = (Actual Yield / Theoretical Yield) * 100% = (26.7 / 31.5) * 100% = 84.8%