Answer
See the answer below.
Work Step by Step
Number of moles of nickel:
$0.450\ g\div 58.693\ g/mol=7.67×10^{-3}\ mol$
Number of moles of CO:
$n=1.50\ L×(418/760×1\ atm)/0.082057\ L.atm/mol.K\div 298\ K$
$n=0.0337\ mol$
Ratio: $0.0337/0.00767=4.4\gt4$
CO is in excess
Theoretical yield:
$7.67×10^{-3}\ mol×170.7\ g/mol=1.31\ g$
