Question

The Mond's process is given below.
$Ni+4CO\to Ni(CO{)}_4$
Calculate the amount of excess reagent if 4 moles of $Ni$,reacts with $CO$, which is produced in the process in which 6 g carbon is mixed with $44gC{O}_2$ and the percentage yield for this reaction is 80 % (molecular mass of $Ni:58u$)

Answer 1

The reaction for the production of $CO$ is: $C+C{O}_2\to 2CO$
Again, 6 g carbon $=\frac{6}{12}\text{mol of carbon}$
$=0.5\text{mol of carbon}$
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, $Ni+4CO\to Ni(CO{)}_4$
So, 4 moles $CO$ reacts with 1.0 moles of $Ni$
Hence, 0.8 moles $CO$ will react with 0.2 moles of $Ni$
Moles of $Ni$ consumed = $0.2mol$

Moles of excess $Ni$ = $4-0.2mol=3.8mol$

Molar mass of $Ni$ = $58g/mol$ (given)
$\therefore$ 3.8 moles of $Ni=220.4g$