If isobutane and n-butane are present in a ga | vedclass

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(A) Both isobutane and $n$-butane are isomers with the same molecular formula,$C_4H_{10}$.The balanced combustion reaction for $C_4H_{10}$ is:$2 C_4H_

Vedclass · Accepted Answer

$35.862$

Vedclass · Answer

(A) Both isobutane and $n$-butane are isomers with the same molecular formula,$C_4H_{10}$.The balanced combustion reaction for $C_4H_{10}$ is:$2 C_4H_{10} + 13 O_2 ightarrow 8 CO_2 + 10 H_2O$From the stoichiometry,$2 \ moles$ of $C_4H_{10}$ require $13 \ moles$ of $O_2$.Molar mass of $C_4H_{10} = (4 	imes 12) + (10 	imes 1) = 58 \ g/mol$.Moles of $10 \ kg$ $(10000 \ g)$ of butane $= \frac{10000}{58} \ moles$.Moles of $O_2$ required $= \frac{13}{2} 	imes \frac{10000}{58} \ moles$.Mass of $O_2$ required $= 	ext{Moles} 	imes 	ext{Molar mass of } O_2 = \frac{13}{2} 	imes \frac{10000}{58} 	imes 32 \ g$.Mass of $O_2 = \frac{13 	imes 5000 	imes 32}{58} \ g = \frac{2080000}{58} \ g \approx 35862 \ g = 35.862 \ kg$.

If isobutane and $n$-butane are present in a gas,then how much oxygen should be required for complete combustion of $10 \ kg$ of this gas? (in $kg$)

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