An ideal gas mixture of C_2H_6 and C_2H_4 occ | vedclass

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(D) $1$. Calculate total moles of the mixture $(n_{total})$ using the ideal gas law: $PV = nRT$. Given $P = 1 \ atm$,$V = 28 \ L$,$R = 0.0821 \ L \ at

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$0.6$

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(D) $1$. Calculate total moles of the mixture $(n_{total})$ using the ideal gas law: $PV = nRT$. Given $P = 1 \ atm$,$V = 28 \ L$,$R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$,and $T = 273 \ K$. $n_{total} = \frac{1 	imes 28}{0.0821 	imes 273} \approx 1.25 \ mol$.$2$. Let $x$ be the moles of $C_2H_6$ and $y$ be the moles of $C_2H_4$. So,$x + y = 1.25$.$3$. Combustion reactions: $C_2H_6 + 3.5O_2 ightarrow 2CO_2 + 3H_2O$$C_2H_4 + 3O_2 ightarrow 2CO_2 + 2H_2O$$4$. Moles of $O_2$ required: $3.5x + 3y = \frac{128}{32} = 4$.$5$. Solve the system of equations: $x + y = 1.25 \Rightarrow x = 1.25 - y$.$6$. Substitute into the second equation: $3.5(1.25 - y) + 3y = 4$ $\Rightarrow 4.375 - 3.5y + 3y = 4$ $\Rightarrow 0.5y = 0.375$ $\Rightarrow y = 0.75$.$7$. Mole fraction of $C_2H_4 = \frac{y}{x+y} = \frac{0.75}{1.25} = 0.6$.

An ideal gas mixture of $C_2H_6$ and $C_2H_4$ occupies a volume of $28 \ L$ at $1 \ atm$ and $273 \ K$. This mixture reacts completely with $128 \ g$ of $O_2$ to produce $CO_2$ and $H_2O_{(l)}$. What is the mole fraction of $C_2H_4$ in the mixture?

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