A person consumes 640 , g of O_2 daily. If al | vedclass

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(C) The combustion reaction of sucrose is: $C_{12}H_{22}O_{11} + 12O_2 \rightarrow 12CO_2 + 11H_2O$.The enthalpy of combustion is $\Delta H = -5645 \,

Vedclass · Accepted Answer

$9408$

Vedclass · Answer

(C) The combustion reaction of sucrose is: $C_{12}H_{22}O_{11} + 12O_2 ightarrow 12CO_2 + 11H_2O$.The enthalpy of combustion is $\Delta H = -5645 \, kJ \, mol^{-1}$.Moles of $O_2$ consumed per day $= \frac{640 \, g}{32 \, g \, mol^{-1}} = 20 \, mol$.From the stoichiometry of the reaction,$12 \, mol$ of $O_2$ is required for $1 \, mol$ of sucrose.Moles of sucrose consumed $= \frac{20}{12} \, mol = 1.667 \, mol$.Mass of sucrose consumed $= \frac{20}{12} 	imes 342 \, g \, mol^{-1} = 570 \, g$.Heat produced $= 	ext{Moles of sucrose} 	imes |\Delta H| = \frac{20}{12} 	imes 5645 \, kJ = 9408.33 \, kJ$.

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