If an LPG cylinder contains a mixture of buta | vedclass

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(D) The combustion reaction for butane $(C_4H_{10})$ is: $C_4H_{10} + \frac{13}{2} O_2 	o 4CO_2 + 5H_2O$ The molar mass of butane is $58 \ g/mol$. Fo

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$3.58$

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(D) The combustion reaction for butane $(C_4H_{10})$ is: $C_4H_{10} + \frac{13}{2} O_2 	o 4CO_2 + 5H_2O$ The molar mass of butane is $58 \ g/mol$. For $1 \ kg$ $(1000 \ g)$ of butane,the number of moles is $n = \frac{1000}{58} \ mol$. According to the stoichiometry,$1 \ mol$ of $C_4H_{10}$ requires $\frac{13}{2} \ mol$ of $O_2$. So,moles of $O_2$ required = $\frac{13}{2} 	imes \frac{1000}{58} \ mol$. Mass of $O_2$ required = $\frac{13}{2} 	imes \frac{1000}{58} 	imes 32 \ g$. Mass of $O_2$ = $3586.2 \ g \approx 3.58 \ kg$.

If an $LPG$ cylinder contains a mixture of butane and isobutane,what is the amount of oxygen required for the combustion of $1 \ kg$ of this mixture in $kg$?

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