Question #b0450

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Answer 1 · 2018-01-27T05:32:18

$2C_4H_10+13O_2=8CO_2+10H_2O$

As per the above balanced equation of the given equation the interacting ratio of number of moles of reactants is

$("moles of "O_2)/("moles of "C_4H_10)=13/2$

Molar mass of $C_4H_10=(4*12+10*1)=58gmol^-1$

So we can write

$("At STP Volume of "O_2)/("mass of "C_4H_10)=(13xx22.4)/(2xx58)Lg^-1$

$=(13xx22.4xx1000)/(2xx58)Lkg^-1$

$~~2510.34Lkg^-1$

So at STP the volume of oxygen needed for complete combustion of 1kg of butane will be

$~~2510.34L$