Question #6d6ed

1 Answer
Nov 7, 2015

Answer:

#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.