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

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_{2}) = pV/RT = (1 x 1.178) / (0.082 x 273) = 0.053 mol

According to chemical equation:

n(C_{4}H_{10}) = n(O_{2}) x 2 / 13 = 0.053 x 2 / 13 = 8.15x 10^-3 mol

m(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) = 0.473g

Answer : m(C_{4}H_{10}) = 0.473g

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

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