Respuesta :
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.
[tex]\text{ Moles of Fe}=\frac{\text{ Mass of Fe}}{\text{ Molar mass of Fe}}=\frac{198.5g}{56g/mole}=3.545moles[/tex]
The balanced chemical reaction is,
[tex]Fe_2O_3(s)+3CO(g)\rightarrow 3CO_2(g)+2Fe(s)[/tex]
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 [tex]\frac{3}{2}\times 3.545=5.3175[/tex] moles of carbon monoxide
Now we have to calculate the mass of carbon monoxide, CO.
[tex]\text{ Mass of CO}=\text{ Moles of CO}\times \text{ Molar mass of CO}[/tex]
[tex]\text{ Mass of CO}=(5.3175moles)\times (28g/mole)=148.89g[/tex]
Therefore, the grams of carbon monoxide needed are 148.89 g
