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Question #6e7c9

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Answer 1 · 2014-12-07T09:59:48

The answers are $V_{H_{2}} = 43.1 L$ $V_(H_2) = 43.1 L$ and $V_{C O} = 21.5 L$ $V_(CO) = 21.5 L$ .

Starting from the ideal gas law $P V = n R T$ $PV = nRT$ and from the chemical reaction's balanced equation

$C O (g) + 2 H_{2} (g) \to C H_{3} O H (g)$ $CO(g) + 2H_2(g) -> CH_3OH(g)$ , one can see that we get a $1 : 1$ $1:1$ mole ratio for $C O$ $CO$ and $C H_{3} O H$ $CH_3OH$ , and a $2 : 1$ $2:1$ mole ratio for $H_{2}$ $H_2$ and $C H_{3} O H$ $CH_3OH$ .

The number of moles of $C H_{3} O H$ $CH_3OH$ can be calculated from

$n_{C H_{3} O H} = \frac{m_{C H_{3} O H}}{32 \frac{g}{m o l}} = \frac{25.8 g}{32 \frac{g}{m o l}} = 0.81$ $n_(CH_3OH) = m_(CH_3OH)/(32 g/(mol)) =(25.8 g)/(32 g/(mol)) = 0.81$ moles.
Subsequently, the number of moles for $C O$ $CO$ and $H_{2}$ $H_2$ are

$n_{C O} = 0.81$ $n_(CO) = 0.81$ moles, and $n_{H_{2}} = 1.62$ $n_(H_2) = 1.62$ moles.

Therefore, the volume of $H_{2}$ $H_2$ required for the reaction to take place is

$V_{H_{2}} = \frac{n_{H_{2}} R T}{P} = \frac{1.62 \cdot 0.082 \cdot 319.15}{\frac{748}{760}} = 43.1$ $V_(H_2) = (n_(H_2)RT)/P = (1.62 * 0.082 * 319.15)/ (748/760) = 43.1$ L ( I've converted pressure into $a t m$ $atm$ anddegrees Celsius into $K$ $K$ ).

The volume of $C O$ $CO$ required will be

$V_{C O} = \frac{n_{C O} R T}{P} = \frac{0.81 \cdot 0.082 \cdot 319.15}{\frac{748}{760}} = 21.5$ $V_(CO) = (n_(CO) RT)/P = (0.81 * 0.082 * 319.15)/(748/760) = 21.5$ L, which is approximately half of the volume of $H_{2}$ $H_2$ determined (half the number of moles -> half the volume);

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Question #6e7c9

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