Answer to Question #253175 in General Chemistry for JmA

Question

How many grams of P2O5 and H2O will be produced in the reaction?

Answer 1

Let use the following example:

If 17.0g of phosphine is mixed with 16.0g of O2 and reaction occurs. How many grams of P2O5 and H2O will be produced in the reaction?

2PH₃ + 4O₂ → P₂O₅ + 3H₂O

M(PH₃) = 34.00 g/mol

n(PH₃) $= \frac{17.0}{34.0} = 0.5 \;mol$

M(O₂) = 32.00 g/mol

n(O₂) $= \frac{16.0}{32.0} = 0.5 \;mol$

According to the reaction equation for one mole of phosphine we need two moles of O2. So, O2 is a limitting reactant.

According to the reaction:

n(P₂O₅) $= \frac{1}{4}n(O_2) = \frac{1}{4} \times 0.5 = 0.125 \;mol$

M(P₂O₅) = 142.0 g/mol

m(P₂O₅) $= 0.125 \times 142.0 = 17.75 \;g$

According to the reaction:

n(H₂O) $= \frac{3}{4}n(O_2) = \frac{3}{4} \times 0.5 = 0.375 \;mol$

M(H₂O) = 18 g/mol

m(H₂O) $= 0.375 \times 18.0 = 6.75 \;g$