Question

Find the amount of nickel carbonyl produced when 60 g of carbon reacts with 242 g of carbon dioxide and 234.8 g of nickel.
$C+C{O}_2\to 2CO$
$Ni+4CO\to Ni(CO{)}_4$
(Molar mass of nickel = 58.7 g/mol)

• A. 426.75 g
• B. 485.60 g
• C. 861.25 g
• D. 551.20 g

Answer 1

The correct option is A 426.75 g

$C+C{O}_2\to 2CONi+4CO\to Ni(CO{)}_4$
$\text{Moles of C =}\frac{\text{given mass}}{\text{molar mass}}=\frac{60}{12}=5mol\text{Moles of}C{O}_2=\frac{\text{given mass}}{\text{molar mass}}=\frac{242}{44}=5.5mol$
Finding the limiting reagent:
$ForC=\frac{\text{given moles}}{\text{stoichiometric coefficient}}=\frac{5}{1}=5mol\text{For}C{O}_2=\frac{\text{given moles}}{\text{stoichiometric coefficient}}=\frac{1.5}{1}=5.5mol$
So, carbon will be the limiting reagent.
1 mol of carbon produces 2 mol of CO.
5 mol will produce 10 mol of CO.
In the second reaction,
$\text{Moles of Ni}=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{234.8}{58.7}=4.0mol$

Finding the limiting reagent:
$ForNi=\frac{\text{given moles}}{\text{stoichiometric coefficient}}=\frac{4}{1}=4mol$
$ForCO=\frac{\text{given moles}}{\text{stoichiometric coefficient}}=\frac{10}{4}=2.5mol$
Limiting reagent will be CO.
1 mole of CO produces 0.25 moles of $Ni(CO{)}_4$
10 moles will produce 2.5 moles of $Ni(CO{)}_4$
Mass of $Ni(CO{)}_4$ produced
= $2.5\times 170.7=426.75g$