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Answer 1 · 2015-11-07T01:07:19

$F e_{2} O_{3}$ $Fe_2O_3$ + $3 H_{2}$ $3H_2$ $\to$ $rarr$ $2 F e$ $2Fe$ + $3 H_{2} O$ $3H_2O$

All you need to do is tally the atoms.

$F e_{2} O_{3}$ $Fe_2O_3$ + $H_{2}$ $H_2$ $\to$ $rarr$ $F e$ $Fe$ + $H_{2} O$ $H_2O$ (unbalanced)

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

Balance the easiest atom first.

$F e_{2} O_{3}$ $Fe_2O_3$ + $H_{2}$ $H_2$ $\to$ $rarr$ $2 F e$ $color (red) 2Fe$ + $H_{2} O$ $H_2O$

left side: Fe = 2; O = 3; H = 2
right side: Fe = (1 x $2$ $color (red) 2$ ) = 2; O = 1; H = 2

$F e_{2} O_{3}$ $Fe_2O_3$ + $H_{2}$ $H_2$ $\to$ $rarr$ $2 F e$ $2Fe$ + $3 H_{2} O$ $color (green) 3H_2O$

left side: Fe = 2; O = 3; H = 2
right side: Fe = (1 x 2) = 2; O = (1 x $3$ $color (green) 3$ ) = 3; H = (2 x $3$ $color (green) 3$ ) = 6

Since $H_{2} O$ $H_2O$ is a substance, you need to also multiply the coefficient 3 to its $H$ $H$ atom.

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

$F e_{2} O_{3}$ $Fe_2O_3$ + $3 H_{2}$ $color (blue)3H_2$ $\to$ $rarr$ $2 F e$ $2Fe$ + $3 H_{2} O$ $3H_2O$

left side: Fe = 2; O = 3; H = (2 x $3$ $color (blue)3$ ) = 6
right side: Fe = (1 x 2) = 2; O = (1 x 3) = 3; H = (2 x 3) = 6

The equation in now balanced.