Balance the equations below by adding coefficients. Enter a number in each blank, entering "1" when necessary.

1. CO2 → C+O2

___CO2 →___ C + ___O2

2. Fe2 + O2 → FeO

___Fe2 + ___O2 →___ FeO

3. Al + CuO → Al2O3 + Cu

___Al + ___CuO →___ Al2O3 + ___Cu

1CO2 → 1C + 1O2

1Fe2 + 1O2 → 2FeO

2Al + 3CuO → 1Al2O3 + 3Cu

Explanation:

1. CO2 → C+O2

On both sides we have 1x C and 2X O.

The equation is balanced

1CO2 → 1C + 1O2

2. Fe2 + O2 → FeO

On the left side we 2x Fe, On the right side we have 1x Fe (in FeO). To balance the amount of Fe on both sides, we have to multiply FeO (on the right) by 2. If we multiply FeO by 2, we'll have 2x O on the right side. This is the same amount on the left side. The equation is balanced

1Fe2 + 1O2 → 2FeO

3. Al + CuO → Al2O3 + Cu

On the left side we have 1x Al, on the right side we have 2x Al. To balance the amount of Al we have to multiply Al (on the left) by 2.

We have 3x O on the right side, on the left side we have 1x O. To balance the amount of O on both sides, we have to multiply CuO (on the left) by 3.

We have 3x Cu on the left side (in 3CuO) and 1x Cu on the right side. To balance the amount of Cu we have to multiply Cu (on the right) by 3.

Now the equation is balanced.

2Al + 3CuO → 1Al2O3 + 3Cu

1. CO₂ → C + O₂

2. Fe₂ + O₂ → 2FeO

3. 2Al + 3CuO→ Al₂O₃ + 3Cu

Explanation:

1. 1 mol of CO₂ decomposes to 1 mol of C and 1 mol of oxygem

2. 1 mol of Fe₂ reacts with 1 mol of oxygen to produce 2 moles of iron (II) oxide

3. 2 moles of Al, reacts with 3 moles of cupper (II) oxide to produce 1 mol of aluminum dioxide and 3 moles of cupper