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(C) Let $n_1$ be the number of moles of $CO_2$ and $n_2$ be the number of moles of $O_2$.From the ideal gas equation, $PV = nRT$, where $n = n_1 + n_2

Vedclass · Accepted Answer

$0.21$ and $0.12$

Vedclass · Answer

(C) Let $n_1$ be the number of moles of $CO_2$ and $n_2$ be the number of moles of $O_2$.From the ideal gas equation, $PV = nRT$, where $n = n_1 + n_2$.Given: $P = 100 	ext{ kPa} = 10^5 	ext{ Pa}$, $V = 8310 	ext{ cm}^3 = 8.31 	imes 10^{-3} 	ext{ m}^3$, $T = 300 	ext{ K}$, $R = 8.31 	ext{ J/mol.K}$.Calculating total moles $n = n_1 + n_2 = \frac{PV}{RT} = \frac{10^5 	imes 8.31 	imes 10^{-3}}{8.31 	imes 300} = \frac{100}{300} = \frac{1}{3} \approx 0.333 	ext{ mol}$.The total mass is $m = M_1 n_1 + M_2 n_2 = 44 n_1 + 32 n_2 = 13.2 	ext{ g}$.We have the system of equations:$1) n_1 + n_2 = 0.333$$2) 44 n_1 + 32 n_2 = 13.2$From $(1)$, $n_2 = 0.333 - n_1$. Substituting into $(2)$:$44 n_1 + 32(0.333 - n_1) = 13.2$$44 n_1 + 10.656 - 32 n_1 = 13.2$$12 n_1 = 2.544$$n_1 = 0.212 \approx 0.21 	ext{ mol}$.$n_2 = 0.333 - 0.212 = 0.121 \approx 0.12 	ext{ mol}$.Thus, the number of moles of $CO_2$ and $O_2$ are $0.21$ and $0.12$ respectively. The correct option is $C$.

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