The Mond's process is given below.
$N i + 4 C O \to N i (C O)_{4}$
Calculate the amount of excess reagent if 4 moles of $N i$ ,reacts with $C O$ , which is produced in the process in which 6 g carbon is mixed with $44 g C O_{2}$ and the percentage yield for this reaction is 80 % (molecular mass of $N i : 58 u$ )

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Solution

The reaction for the production of $C O$ is: $C + C O_{2} \to 2 C O$
Again, 6 g carbon $= \frac{6}{12} mol of carbon$
$= 0.5 mol of carbon$
44g $C O_{2} = \frac{44}{44}$ mol of $C O_{2}$
= 1 mol of $C O_{2}$

So, carbon is the limiting reagent here.

Moles of CO produced, $= 0.5 \times 2 \times 0.8 = 0.8$ mol
(since the yield is 80 %)

Again, $N i + 4 C O \to N i (C O)_{4}$
So, 4 moles $C O$ reacts with 1.0 moles of $N i$
Hence, 0.8 moles $C O$ will react with 0.2 moles of $N i$
Moles of $N i$ consumed = $0.2 m o l$

Moles of excess $N i$ = $4 - 0.2 m o l = 3.8 m o l$

Molar mass of $N i$ = $58 g / m o l$ (given)
$∴$ 3.8 moles of $N i = 220.4 g$

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