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Question

Calculate the amount of Ni needed in the Mond's process given below

Ni + 4CO Ni(CO)4

If CO used in this process is obtained through a process, in which 6 g of carbon is mixed with 44 g CO2 .

Answer 1

Solution

(1)

C + CO2 $\longrightarrow$ 2CO

Given moles $\left( \frac{6}{12} \right)$ = 0.5 $\left( \frac{44}{44} \right)$ = 1

So C is limiting reagent

$\therefore$ CO formed = 1 moles

Now moles of Ni need to react with 1 moles of CO are

$\frac{1}{4}$ × 58.7 = 14.675 gm.