Answer: [tex]CO[/tex] is the limiting reagent and [tex]Fe_2O_3[/tex] is the excess reagent.
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} Fe_2O_3=\frac{42.7g}{159.69g/mol}=0.267moles[/tex]
[tex]\text{Moles of} CO=\frac{13.2g}{28g/mol}=0.471moles[/tex]
The given balanced equation is :
[tex]Fe_2O_3(s)+3CO(g)\rightarrow 2Fe(s)+3CO_2(g)[/tex]
According to stoichiometry :
3 moles of [tex]CO[/tex] require = 1 mole of [tex]Fe_2O_3[/tex]
Thus 0.471 moles of [tex]CO[/tex] will require=[tex]\frac{1}{3}\times 0.471=0.157moles[/tex] of [tex]Fe_2O_3[/tex]
As given amount of [tex]Fe_2O_3[/tex] is more than the required amount , it is the excess reagent.Thus [tex]CO[/tex] is the limiting reagent as it limits the formation of product.