Question

Question #6d6ed

Answer 1

`Fe_2O_3` + `3H_2` `rarr` `2Fe` + `3H_2O`

Explanation:

All you need to do is tally the atoms.

`Fe_2O_3` + `H_2` `rarr` `Fe` + `H_2O` (unbalanced)

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

Balance the easiest atom first.

`Fe_2O_3` + `H_2` `rarr` `color (red) 2Fe` + `H_2O`

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

`Fe_2O_3` + `H_2` `rarr` `2Fe` + `color (green) 3H_2O`

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

Since `H_2O` 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.

`Fe_2O_3` + `color (blue)3H_2` `rarr` `2Fe` + `3H_2O`

left side: Fe = 2; O = 3; H = (2 x `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.