Question #21c1a

1 Answer
Junaid Mirza
Nov 26, 2017

#"1737.5 L"# of air

Explanation:

Reaction equation of combustion of butane:

#C_4H_10 + 13/2O_2 -> 4CO_2 + 5H_2O#

From above equation, At STP #"22.4 L"# of butane requires #13/2xx22.4L# of oxygen gas.

So, 56 litres of butane gas require

#(56 × 13/2 × cancel"22.4 L")/cancel"22.4 L" = "364 L"# of Oxygen gas.

By volume, air contains #20.95%# Oxygen.

Volume of air#= ("364 L" × 100)/20.95 = "1737.5 L"#

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