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.

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. (A) 14.675 g (B) 29 g (C) 58 g (D) 28 g

Tardigrade Admin · Accepted Answer

underset textGiven moles underset((6/12)) = 0.5C + underset((44/44)) = 1CO2 → 2CO So C is limiting reagent ∴ CO formed = 1 mole Now moles of Ni need to react with 1 mole of CO are (1/4) × 58.7 = 14.675 g

Q. Calculate the amount of $N i$ needed in the Mond’s process given below
$N i + 4 CO \to N i (CO)_{4}$
If $CO$ used in this process is obtained through a process, in which $6 g$ of carbon is mixed with $44 g C O_{2}$ .

$14.675 g$

$29 g$

$58 g$

$28 g$

$Given moles (\frac{6}{12}) = 0.5 C + (\frac{44}{44}) = 1 C O_{2} \to 2 CO$
So $C$ is limiting reagent
$∴ CO$ formed $= 1$ mole
Now moles of $N i$ need to react with $1$ mole of $CO$ are
$\frac{1}{4} \times 58.7 = 14.675 g$

Q. 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 6g of carbon is mixed with 44gCO2​.

14.675g

29g

58g

28g