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Limiting Reagent or Reactant

Carbon monoxi...

Question

Carbon monoxide ( $C O$ ) and hydrogen ( $H_{2}$ ) react to form methanol $(C H_{3} O H)$ according to the following reaction:
$C O (g) + 2 H_{2} (g) ⟶ C H_{3} O H (l)$
How much $C H_{3} O H (l)$ (in mg) is obtained from $0.01$ mol of $C O$ and $0.08$ g $H_{2}$ g?

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Solution

$C O (g) + 2 H_{2} (g) ⟶ C H_{3} O H (l)$
$1 m o l 2 m o l 1 m o l$
$0.01 m o l 0.02 m o l 0.01 m o l$
$0.02 m o l 0.04 m o l 0.02 m o l$
$C O (g) = 0.01 m o l$
$H_{2} (g) = 0.08 g = 0.04 m o l$
Thus, $C O (g)$ is limiting agent.
$C H_{3} O H (l) f o r m e d = 0.01 m o l = 0.32 g = 320 m g$ .

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