A student prepared methanol in the laboratory by the addition of $266 g$ of $C O$ and $60 g$ of $H_{2}$ , as shown below.
$C O (g) + H_{2} (g) \to C H_{3} O H (l)$
(a) Find out the limiting reagent.
(b) Calculate the amount of methanol produced and the amount of excess reactant remaining after the completion of the reaction.

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Solution

$C O_{(g)} + {H_{2}}_{(g)} ⟶ {C H_{3} O H}_{(l)}$

$1$ $m o l e$ $C O$ reacts with $1$ $m o l e$ of $H_{2}$ .

$∴ 28$ $g$ $C O$ reacts with $2$ $g$ of $H_{2}$

$∴ 266$ $g$ $C O$ reacts with $\frac{266 \times 2}{28} = 19$ $g$ of $H_{2}$

But $60$ $g$ $H_{2}$ is present.

Therefore, $C O$ is the limiting reagent.

$∴ H_{2}$ left $= 60 - 19 = 41$ $g$

Moles of $C O$ related $=$ Moles of $C H_{3} O H$ formed.

$∴$ Amount of method produced $= 32 \times \frac{266}{28} = 304$ $g$

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A student prepared methanol in the laboratory by the addition of 266 g of CO and 60 g of H2, as shown below.CO(g)+H2(g)→CH3OH(l)(a) Find out the limiting reagent.(b) Calculate the amount of methanol produced and the amount of excess reactant remaining after the completion of the reaction.