# 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

2C

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(O

According to chemical equation:

n(C

m(C

Answer : m(C

2C

_{4}H_{10}+ 13O_{2}= 8CO_{2}+ 10H_{2}OAccording 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(O

_{2}) = pV/RT = (1 x 1.178) / (0.082 x 273) = 0.053 molAccording to chemical equation:

n(C

_{4}H_{10}) = n(O_{2}) x 2 / 13 = 0.053 x 2 / 13 = 8.15x 10^-3 molm(C

_{4}H_{10}) = n(C_{4}H_{10}) x M_{W}(C_{4}H_{10}) = n(C_{4}H_{10}) x (4 xM_{W}(C) + 10 xM_{W}(H)) = 8.15 x 10^-3 x (4 x 12 + 10 x 1) = 0.473gAnswer : m(C

_{4}H_{10}) = 0.473g
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