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 35.0 grams of PH3?
c. How many grams of P2O5 and H2O will be produced in the reaction?
d. If the reaction give 64% yield. How many grams of PH3 should be used to give an actual yield of 250 grams of P2O5?

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) $= 2.74 \times 40 = 109.6 \;g$