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Answer to Question #24952 in Other Chemistry for jordan

Question #24952
At STP, how many grams of butane would be needed to react with 1178.0 mL of oxygen gas?
Expert's answer
The chemical equation for the burning reaction is
2C4H10 + 13O2 = 8CO2 + 10H2O
According to the ideal gas law
pV = nRT
n = pV/RT
STP:
p = 1 atm
V = 22.4 L
T = 273 K
R = 0.082 atm x L/ (mol x K)

n(O2) = pV/RT = (1 x 1.178) / (0.082 x 273) = 0.053 mol

According to chemical equation:
n(C4H10) = n(O2) x 2 / 13 = 0.053 x 2 / 13 = 8.15x 10^-3 mol
m(C4H10) = n(C4H10) x MW(C4H10) = n(C4H10) x (4 xMW(C) + 10 xMW(H)) = 8.15 x 10^-3 x (4 x 12 + 10 x 1) = 0.473g

Answer : m(C4H10) = 0.473g

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