Question

The Mond process produces pure nickel metal via the thermal decomposition of nickel tetracarbonyl: Ni(CO)4 (l) → Ni (s) + 4CO (g). What volume (L) of CO is formed from the complete decomposition of 444 g of Ni(CO)4 at 752 torr and 22.0 °C?

Answer 1

Answer: The volume of CO formed is 254.43 L.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$  

Given mass of $Ni(CO)_4$ = 444 g

Molar mass of $Ni(CO)_4$ = 170.73 g/mol

Putting values in above equation, we get:

$\text{Moles of }Ni(CO)_4=\frac{444g}{170.73g/mol}=2.60mol$

For the given chemical reaction:

$Ni(CO)_4(l)\rightarrow Ni(s)+4CO(g)$

By stoichiometry of the reaction:

1 mole of nickel tetracarbonyl produces 4 moles of carbon monoxide.

So, 2.60 moles of nickel tetracarbonyl will produce = $\frac{4}{1}\times 2.60=10.4mol$ of carbon monoxide.

Now, to calculate the volume of the gas, we use ideal gas equation, which is:

PV = nRT

where,

P = Pressure of the gas = 752 torr

V = Volume of the gas = ? L

n = Number of moles of gas = 10.4 mol

R = Gas constant = $62.364\text{ L Torr }mol^{-1}K^{-1}$

T = Temperature of the gas = $22^oC=(273+22)K=295K$

Putting values in above equation, we get:

$752torr\times V=10.4mol\times 62.364\text{ L Torr }mol^{-1}K^{-1}\times 295K\\\\V=254.43L$

Hence, the volume of CO formed is 254.43 L.