In a 1 L vessel, 10 text{moles} of methane | vedclass

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(B) The combustion reaction for methane is as follows:$CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$Initial moles:$CH_4 = 10, O_2 = 50, CO_2 = 0, H_2O = 0$Sin

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$30, 20, 10$

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(B) The combustion reaction for methane is as follows:$CH_4 + 2O_2 ightarrow CO_2 + 2H_2O$Initial moles:$CH_4 = 10, O_2 = 50, CO_2 = 0, H_2O = 0$Since $CH_4$ is the limiting reagent, it will be completely consumed.For $10 \ 	ext{moles}$ of $CH_4$, $2 	imes 10 = 20 \ 	ext{moles}$ of $O_2$ are consumed.Final moles:$O_2 = 50 - 20 = 30 \ 	ext{moles}$$CO_2 = 10 \ 	ext{moles}$$H_2O = 2 	imes 10 = 20 \ 	ext{moles}$Thus, the number of moles of $O_2$, $H_2O$, and $CO_2$ are $30, 20, 10$ respectively.Option $B$ is the correct answer.

In a $1 \ L$ vessel, $10 \ \text{moles}$ of methane and $50 \ \text{moles}$ of $O_2$ are present. The number of moles of $O_2$, water, and $CO_2$ present in the vessel after the vessel is heated to burn methane completely are respectively:

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