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(A) The balanced chemical equation for the combustion of ethane $(C_2H_6)$ is: $2C_2H_6(g) + 7O_2(g) \rightarrow 4CO_2(g) + 6H_2O(g)$ Calculate the mo

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$4.25$

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(A) The balanced chemical equation for the combustion of ethane $(C_2H_6)$ is: $2C_2H_6(g) + 7O_2(g) ightarrow 4CO_2(g) + 6H_2O(g)$ Calculate the moles of reactants: Molar mass of $C_2H_6 = 30 \ g/mol$. Moles of $C_2H_6 = 15 \ g / 30 \ g/mol = 0.5 \ mol$. Molar mass of $O_2 = 32 \ g/mol$. Moles of $O_2 = 112 \ g / 32 \ g/mol = 3.5 \ mol$. Determine the limiting reagent: For $0.5 \ mol$ of $C_2H_6$,required $O_2 = (7/2) 	imes 0.5 = 1.75 \ mol$. Since we have $3.5 \ mol$ of $O_2$,$C_2H_6$ is the limiting reagent. Calculate moles of products and remaining reactants: $C_2H_6$ consumed = $0.5 \ mol$ (Remaining = $0 \ mol$). $O_2$ consumed = $1.75 \ mol$ (Remaining = $3.5 - 1.75 = 1.75 \ mol$). $CO_2$ produced = $(4/2) 	imes 0.5 = 1.0 \ mol$. $H_2O$ produced = $(6/2) 	imes 0.5 = 1.5 \ mol$. Total moles of gaseous substances = $n(O_2) + n(CO_2) + n(H_2O) = 1.75 + 1.0 + 1.5 = 4.25 \ mol$.

Complete combustion of ethane gives only gaseous products. In a closed vessel,$15 \ g$ of ethane and $112 \ g$ of $O_2$ were allowed to completely react. What is the total number of moles of gaseous substances present in the vessel at the end of the reaction?

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