Consider the complete combustion of butane. T | vedclass

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(D) The balanced chemical equation for the combustion of butane is:$C_{4}H_{10} + \frac{13}{2} O_{2} \longrightarrow 4 CO_{2} + 5 H_{2}O$Calculate the

Vedclass · Accepted Answer

$464$

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(D) The balanced chemical equation for the combustion of butane is:$C_{4}H_{10} + \frac{13}{2} O_{2} \longrightarrow 4 CO_{2} + 5 H_{2}O$Calculate the moles of water produced:$	ext{Moles of } H_{2}O = \frac{72.0 \ g}{18.0 \ g/mol} = 4.0 \ mol$From the stoichiometry of the reaction,$1 \ mol$ of $C_{4}H_{10}$ produces $5 \ mol$ of $H_{2}O$.Therefore,the moles of $C_{4}H_{10}$ required to produce $4.0 \ mol$ of $H_{2}O$ is:$	ext{Moles of } C_{4}H_{10} = \frac{1}{5} 	imes 4.0 = 0.8 \ mol$The molar mass of butane $(C_{4}H_{10})$ is $(4 	imes 12) + (10 	imes 1) = 58 \ g/mol$.Calculate the mass of butane:$	ext{Mass of } C_{4}H_{10} = 0.8 \ mol 	imes 58 \ g/mol = 46.4 \ g$Expressing $46.4 \ g$ in the form $.... 	imes 10^{-1} \ g$:$46.4 \ g = 464 	imes 10^{-1} \ g$Thus,the nearest integer value is $464$.

Consider the complete combustion of butane. The amount of butane utilized to produce $72.0 \ g$ of water is $.... \times 10^{-1} \ g$ (in nearest integer).

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