## Chemistry and Chemical Reactivity (9th Edition)

$x=4.0$
Atomic weights (g/mol): $Ni: 58.6934,\ C: 12.011, \ O: 15.999$ For the 0.125 g of Ni $0.125\ g\div 58.6934g/mol=2.13\times10^{-3} mol$ The $Ni(CO)_x$ must have the same amount of moles of Ni: $0.364\ g=2.13\times10^{-3} mol\times[58.6934+x\times(12.011+15.999)]$ Solving for x: $x=4.0$