Answer:
The yield percent is 77.68%
Explanation:
[tex]Fe_{2}O_{3}+3CO\longrightarrow2Fe+3CO_{2}[/tex]
1 mole of [tex]Fe_{2}O_{3}[/tex] yields 2 moles of Fe.
So, 11.2 moles of Iron(III) oxide must yield 22.4 moles of Fe theoretically. This is according to the balanced equation.
Theoretical Yield = 22.4 moles of Fe
Experimental Yield = 17.4 moles of Fe
[tex]Yield\:Percentage=\frac{Experimental\:Yield}{Theoretical\:Yield}\times100\\\\Yield\:Percentage=\frac{17.4}{22.4}\times100=77.68\%[/tex]
Therefore, the yield percent is 77.68%