Answer to Question #251426 in General Chemistry for Jam

Question #251426
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?
1
Expert's answer
2021-10-15T02:55:00-0400

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"

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