Question #6d6ed

Nov 7, 2015

$F {e}_{2} {O}_{3}$ + $3 {H}_{2}$ $\rightarrow$ $2 F e$ + $3 {H}_{2} O$

Explanation:

All you need to do is tally the atoms.

$F {e}_{2} {O}_{3}$ + ${H}_{2}$ $\rightarrow$ $F e$ + ${H}_{2} O$ (unbalanced)

left side: Fe = 2; O = 3; H = 2
right side: Fe = 1; O = 1; H = 2

Balance the easiest atom first.

$F {e}_{2} {O}_{3}$ + ${H}_{2}$ $\rightarrow$ $\textcolor{red}{2} F e$ + ${H}_{2} O$

left side: Fe = 2; O = 3; H = 2
right side: Fe = (1 x $\textcolor{red}{2}$) = 2; O = 1; H = 2

$F {e}_{2} {O}_{3}$ + ${H}_{2}$ $\rightarrow$ $2 F e$ + $\textcolor{g r e e n}{3} {H}_{2} O$

left side: Fe = 2; O = 3; H = 2
right side: Fe = (1 x 2) = 2; O = (1 x $\textcolor{g r e e n}{3}$) = 3; H = (2 x $\textcolor{g r e e n}{3}$) = 6

Since ${H}_{2} O$ is a substance, you need to also multiply the coefficient 3 to its $H$ atom.

Now the only thing left to balance is the $H$ atom on the left.

$F {e}_{2} {O}_{3}$ + $\textcolor{b l u e}{3} {H}_{2}$ $\rightarrow$ $2 F e$ + $3 {H}_{2} O$

left side: Fe = 2; O = 3; H = (2 x $\textcolor{b l u e}{3}$) = 6
right side: Fe = (1 x 2) = 2; O = (1 x 3) = 3; H = (2 x 3) = 6

The equation in now balanced.