Question

$Ni+4CO\to Ni(CO{)}_4$

Calculate the amount of Ni needed in the Mond's process given below if CO used in this process is obtained through a process, in which 6 g of carbon is mixed with 44 g $C{O}_2$.

• A. 14.675 g
• B. 29 g
• C. 58 g
• D. 28 g

Answer 1

The correct option is A 14.675 g

$C+C{O}_2\to 2CO$

1 mole of C reacts with 1 mole of $C{O}_2$, .i.e., 12 g of $C$ reacts with 44 g of $C{O}_2$ but $C$ is present in less amount and hence, $C$ is the limiting reagent.

12 g C forms $2\times 28$ g $CO$ but 6 g form $\frac{(6\times 2\times 28)}{12}$, i.e., 28 g of $CO$.

Number of moles of CO = $\frac{28}{28}$$=1$

4$\times$28 g of $CO$ combine with 58.7 g of $Ni$.

So, 28 g of $CO$ will combine with $\frac{58.7}{4}$ $=14.675$ g of $Ni$.

Hence, the correct option is $A$