Question

Calculate the amount of $Ni$ (in gram) needed in the Mond's process given below:
$Ni+4CO\to Ni(CO{)}_4$
If $CO$ used in this process is obtained through a process, in whcih $6g$ of carbon is mixed with $44g$ $C{O}_2$.
(Given: Molar mass of $Ni$ is 58.69 $g/mol$)

• A. 14.67g
• B. 29.56 g
• C. 58.22 g
• D. 33.62 g

Answer 1

The correct option is A 14.67g

Formation of $CO$ is given as:
$C+C{O}_2\leftrightharpoons 2CO$
given $6g$ of $C$ and $44g$ of $C{O}_2$ are taken
$\therefore$ moles of $C$ are $6/12=0.5$
and moles of $C{O}_2$ are $44/44=1$
So, $C$ is the limiting reagent, $\therefore$ $CO$ formed is 1 mole.
$Ni+4CO\leftrightharpoons Ni(CO{)}_4$
In the above process, 4 moles of $CO$ required 1 mole of $Ni$, so 1 mole of $CO$ will require 1/4 moles of $Ni$.
So weight of $Ni$ will be $=58.69/4=14.67g$