A hydrocarbon containing C and H has 92.3 % C | vedclass

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(A) $1$. Determine the empirical formula: $C = 92.3 \%, H = 7.7 \%$. Moles of $C = \frac{92.3}{12} = 7.69$,Moles of $H = \frac{7.7}{1} = 7.7$. Ratio $

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$120$

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(A) $1$. Determine the empirical formula: $C = 92.3 \%, H = 7.7 \%$. Moles of $C = \frac{92.3}{12} = 7.69$,Moles of $H = \frac{7.7}{1} = 7.7$. Ratio $C:H = 1:1$. Empirical formula is $CH$. $2$. Calculate moles of hydrocarbon: Given mass $= 39 \ g$. Molar mass of $CH = 13 \ g/mol$. Moles of $CH = \frac{39}{13} = 3 \ mol$. $3$. Determine moles of water $(x)$: Reaction: $2H_2O + 2Na ightarrow 2NaOH + H_2$. $2 \ mol$ of $H_2O$ produces $1 \ mol$ of $H_2$. To produce $0.75 \ mol$ of $H_2$,we need $x = 0.75 	imes 2 = 1.5 \ mol$ of $H_2O$. $4$. Combustion reaction: For $CH$ (or $C_2H_2$ unit),the combustion is $C_2H_2 + 2.5 O_2 ightarrow 2 CO_2 + H_2O$. Since we have $3 \ mol$ of $CH$ (which is $1.5 \ mol$ of $C_2H_2$),the reaction is: $1.5 C_2H_2 + 3.75 O_2 ightarrow 3 CO_2 + 1.5 H_2O$. $5$. Calculate mass of $O_2$: Moles of $O_2 = 3.75 \ mol$. Mass of $O_2 = 3.75 	imes 32 = 120 \ g$.

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