# Question b7deb

Nov 7, 2015

${m}_{F e} = 699 g$

#### Explanation:

The reaction is:

$F {e}_{2} {O}_{3} \left(s\right) + 3 C O \left(g\right) \to 2 F e \left(s\right) + 3 C {O}_{2} \left(g\right)$

Since $C O$ is used in excess, therefore, the limiting reactant is $F {e}_{2} {O}_{3}$.

?g Fe(s) = underbrace(1.00xx10^3cancel(gFe_2O_3)xx(1cancel(molFe_2O_3))/(160 cancel(gFe_2O_3)))_(color(blue)("From g to mol"))xxunderbrace((2cancel(molFe))/(1cancel(molFe_2O_3)))_(color(blue)("Molar Ratio"))xxunderbrace((55.9gFe)/(1cancel(molFe)))_(color(blue)("From mol to g"))=color(green)(699gFe)#

I used 3 significant figures since $1.00 k g$ is 3 significant figures.

Explaining the dimensional analysis used above:

The first part:
$1.00 \times {10}^{3} \cancel{g F {e}_{2} {O}_{3}} \times \frac{1 \cancel{m o l F {e}_{2} {O}_{3}}}{160 \cancel{g F {e}_{2} {O}_{3}}}$
is used to convert $g F {e}_{2} {O}_{3}$ to $m o l F {e}_{2} {O}_{3}$

The second part:
$\frac{2 \cancel{m o l F e}}{1 \cancel{m o l F {e}_{2} {O}_{3}}}$
is the molar ration between $F e$ and $F {e}_{2} {O}_{3}$. This is taking from the coefficient of the balanced equation.

The third part:
$\frac{55.9 g F e}{1 \cancel{m o l F e}}$
is used to convert $m o l F e$ to $g F e$.