In a closed vessel at STP,50 , L of CH_4 is i | vedclass

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(B) The combustion reaction is: $CH_4(g) + 2O_2(g) ightarrow CO_2(g) + 2H_2O(l)$.Volume of $O_2$ present in $750 \, L$ of air $= \frac{20}{100} 	im

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$2.2$

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(B) The combustion reaction is: $CH_4(g) + 2O_2(g) ightarrow CO_2(g) + 2H_2O(l)$.Volume of $O_2$ present in $750 \, L$ of air $= \frac{20}{100} 	imes 750 \, L = 150 \, L$.According to the stoichiometry,$1 \, L$ of $CH_4$ requires $2 \, L$ of $O_2$.Therefore,$50 \, L$ of $CH_4$ requires $50 	imes 2 = 100 \, L$ of $O_2$.Remaining volume of $O_2 = 150 \, L - 100 \, L = 50 \, L$.Since $22.4 \, L$ of any gas at $STP$ corresponds to $1 \, mole$,the number of moles of $O_2$ remaining is $\frac{50}{22.4} \approx 2.23 \, moles$.The value closest to this is $2.2 \, moles$.

In a closed vessel at $STP$,$50 \, L$ of $CH_4$ is ignited with $750 \, L$ of air (containing $20 \% \ O_2$). The number of moles of $O_2$ remaining in the vessel on cooling to room temperature is closest to

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