Question

The oxidation of phosphine (PH3) to phosphorus pentoxide (P2O5) is given by the chemical reaction.
__PH3 +__O2 ---> __P2O5 + __H2O
a. Balance the equation
b. How many grams of O2 will react completely with 49.0 grams of PH3?
c. How many grams of P2O5 and H2O will be produced in the reaction?
d. If the actual yield during the experiment is 94 g of p2o5, what is the percent yield of the reaction?

Answer 1

a. 2PH₃ + 4O₂ → P₂O₅ + 3H₂O

b. M(PH₃) = 40 g/mol

M(O₂) = 32 g/mol

$n = \frac{m}{M}$

n(PH₃) = 35/40 = 0.875 mole

n(O₂) = 2n(PH₃) $= 2\times 0.875 = 1.75 \;mol$

$m = n\times M$

m(O₂) $= 1.75 \times 32 = 56\; g$

c. M(P₂O₅) = 284 g/mol

m(H₂O) = 18 g/mol

n(P₂O₅) = 1/2n(PH₃) $= 1/2\times 0.875 = 0.4375\; mol$

m(P₂O₅) = $0.4375 \times 284 = 124.25\; g$

n(H₂O) = 3n(P₂O₅) $= 3\times 0.4375 = 1.315\; mol$

m(H₂O) $= 1.315 \times 18 = 23.67 \;g$

d. Proportion:

250 g — 64 %

x — 100 %

x = 391 g

n(P₂O₅) $= \frac{91}{284} = 1.37 \;mol$

n(PH₃) = 2n(P₂O₅) $= 2 \times 1.37 = 2.74 \;mol$

m(PH₃) $= 2.74 \times 40 = 109.6 \;g$