Question

Consider the balanced chemical reaction below and determine the percent yield for iron if 11.2 moles of Iron(III) oxide yielded 17.4 moles of iron
Fe2O3+3CO>2Fe+3CO2

Answer 1

Answer:

The yield percent is 77.68%

Explanation:

$Fe_{2}O_{3}+3CO\longrightarrow2Fe+3CO_{2}$

1 mole of $Fe_{2}O_{3}$ 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

$Yield\:Percentage=\frac{Experimental\:Yield}{Theoretical\:Yield}\times100\\\\Yield\:Percentage=\frac{17.4}{22.4}\times100=77.68\%$

Therefore, the yield percent is 77.68%