If the heat of combustion of methane is -100 | vedclass

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(A) The combustion reaction of methane is: $CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$.Given,heat of combustion of $CH_4 = -100 \ kJ/mol$.Molar

Vedclass · Accepted Answer

$50$

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(A) The combustion reaction of methane is: $CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$.Given,heat of combustion of $CH_4 = -100 \ kJ/mol$.Molar mass of $CH_4 = 16 \ g/mol$ and $O_2 = 32 \ g/mol$.Number of moles of $CH_4 = \frac{32 \ g}{16 \ g/mol} = 2 \ mol$.Number of moles of $O_2 = \frac{32 \ g}{32 \ g/mol} = 1 \ mol$.According to the stoichiometry,$1 \ mol$ of $CH_4$ requires $2 \ mol$ of $O_2$. Since we have only $1 \ mol$ of $O_2$,$O_2$ is the limiting reagent.The heat evolved depends on the limiting reagent $(O_2)$.For $2 \ mol$ of $O_2$,heat evolved is $100 \ kJ$.For $1 \ mol$ of $O_2$,heat evolved = $\frac{100 \ kJ}{2} = 50 \ kJ$.

If the heat of combustion of methane is $-100 \ kJ/mol$,then calculate the heat evolved when $32 \ g$ of methane reacts with $32 \ g$ of oxygen. (in $kJ$)

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