Question #a7f87

1 Answer
Jun 14, 2017

Answer:

The coefficients in front of the compounds are 1, 3, 2, and 3, respectively so that #Fe_2O_3(s) + 3CO(g) -> 2Fe(s) + 3CO_2(g)#.

Explanation:

When you balance an equation, you want an equal number of atoms on both sides. Let's begin with #Fe#, or iron.

#Fe_2O_3(s) + CO(g) -> Fe(s) + CO_2(g)#

There are 2 atoms of #Fe# on the left but only 1 on the right. Add a coefficient of 2 to the #Fe# atom (or multiply #Fe# by 2) on the right to get 2 #Fe# atoms on the right side of the equation:

#Fe_2O_3(s) + CO(g) -> 2Fe(s) + CO_2(g)#

Now let's focus on #O# atoms since they are next in the equation after #Fe#. There are 4 atoms of O on the left side of the equation, while only 2 on the right side, so add the coefficient 2 to the #O# on the right side (Note: this changes the number of C atoms as well):

#Fe_2O_3(s) + CO(g) -> 2Fe(s) + 2CO_2(g)#

We need to fix the number of #C# atoms on the left now, so add 2 to the #CO# to make the number of Carbons equal (This changes the number of O atoms too):

#Fe_2O_3(s) + 2CO(g) -> 2Fe(s) + 2CO_2(g)#

There are now 5 #O# atoms on the left with 4 on the right, and we can't add a fraction as a coefficient. This means we have to change the numbers so that the numbers of #C# atoms are equal on both sides of the equation. Let's change the coefficients of the compounds with #C# to 3:

#Fe_2O_3(s) + 3CO(g) -> 2Fe(s) + 3CO_2(g)#

There we go! All of the atoms on both sides are now balanced.