Answer : The grams of carbon monoxide needed are 148.89 g
Solution : Given,
Mass of iron, Fe = 198.5 g
Molar mass of iron, Fe = 56 g/mole
Molar mass of carbon monoxide, CO = 28 g/mole
First we have to calculate the moles of iron, Fe.

The balanced chemical reaction is,

From the balanced reaction, we conclude that
2 moles of iron produces from the 3 moles of carbon monoxide
3.545 moles of iron produces from the
moles of carbon monoxide
Now we have to calculate the mass of carbon monoxide, CO.


Therefore, the grams of carbon monoxide needed are 148.89 g