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.
![\text{ Moles of Fe}=\frac{\text{ Mass of Fe}}{\text{ Molar mass of Fe}}=\frac{198.5g}{56g/mole}=3.545moles](https://tex.z-dn.net/?f=%5Ctext%7B%20Moles%20of%20Fe%7D%3D%5Cfrac%7B%5Ctext%7B%20Mass%20of%20Fe%7D%7D%7B%5Ctext%7B%20Molar%20mass%20of%20Fe%7D%7D%3D%5Cfrac%7B198.5g%7D%7B56g%2Fmole%7D%3D3.545moles)
The balanced chemical reaction is,
![Fe_2O_3(s)+3CO(g)\rightarrow 3CO_2(g)+2Fe(s)](https://tex.z-dn.net/?f=Fe_2O_3%28s%29%2B3CO%28g%29%5Crightarrow%203CO_2%28g%29%2B2Fe%28s%29)
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.
![\text{ Mass of CO}=\text{ Moles of CO}\times \text{ Molar mass of CO}](https://tex.z-dn.net/?f=%5Ctext%7B%20Mass%20of%20CO%7D%3D%5Ctext%7B%20Moles%20of%20CO%7D%5Ctimes%20%5Ctext%7B%20Molar%20mass%20of%20CO%7D)
![\text{ Mass of CO}=(5.3175moles)\times (28g/mole)=148.89g](https://tex.z-dn.net/?f=%5Ctext%7B%20Mass%20of%20CO%7D%3D%285.3175moles%29%5Ctimes%20%2828g%2Fmole%29%3D148.89g)
Therefore, the grams of carbon monoxide needed are 148.89 g