Question

Find the amount of carbon needed to produce $683\text{g}$ of $Ni(CO{)}_4$.
$C+C{O}_2\to CO\text{(Yield of reaction = 70 \%)}$
$Ni+4CO\to Ni(CO{)}_4\text{(Yield of Mond’s process = 20 \%)}$
(Molar mass of Nickel = $58.7\text{g/mol}$)

• A. $685.91\text{g}$
• B. $557.25\text{g}$
• C. $748.26\text{g}$
• D. $488.42\text{g}$

Answer 1

The correct option is A $685.91\text{g}$

$C+C{O}_2\to 2CO\text{(i)}$
$Ni+4CO\to Ni(CO{)}_4\text{(ii)}$
In reaction (ii),
1 mol of $Ni(CO{)}_4$ is produced from 4 mol of $CO$
$170.7\text{g}$ of $Ni(CO{)}_4$ is produced from $(4\times 28)\text{g}$ of $CO$
$683\text{g}$ of $Ni(CO{)}_4$ will be obtained from $=\frac{4\times 28}{170.7}\times 683=448.13\text{g}$ of $CO$
Yield of the reaction is given as 20 % Actual amount of CO required $=\frac{448.13}{0.2}=2240.65\text{g}$
In (i) reaction,
2 mol of $CO$ is produced by 1 mol of carbon
$(2\times 28\text{g})$ of $CO$ is produced by $12\text{g}$ of carbon
$2240.65\text{g}$ will be produced by $\frac{12}{2\times 28}\times 2240.65=480.14\text{g}$
Since, the yield of the reaction is given as 70 %
Actual amount of carbon required $=480.14/0.7=685.91\text{g}$