Balance the equation? al2 (so4) 3+? naoh →? al (oh) 3+? na2so4, using the smallest possible integers. what is the sum of the coefficients in the balanced equation?

Answer: 12

Justification:

1) To balance the atoms you can start by Al:

Since there are 2 Al on the left side, add a 2 coefficient in front to Al (OH) ₃

That will lead to this transitorye equation:

Al₂ (SO₄) ₃ + ? NaOH → 2 Al (OH) ₃ + ? Na₂SO₄

2) To balance the three SO₄ radicals on the left, add a 3 in front to Na₂SO₄ on the right, leading to:

Al₂ (SO₄) ₃ + ? NaOH → 2 Al (OH) ₃ + 3 Na₂SO₄

3) To balance the six Na atoms on the right, add a 3 in front to NaOH on the left, leading to

Al₂ (SO₄) ₃ + 6NaOH → 2 Al (OH) ₃ + 3 Na₂SO₄

4) Check that all the species are balanced:

Al: 2 on the left and 2 on the right

SO₄: 3 on the left and 3 on the right

Na: 6 on the left and 6 on the right

OH: 6 on the left and 6 on the right.

Then, the equation is balanced.

5) The sum of the coefficients is 1 + 6 = 7 in the left and 2 + 3 = 5 on the right. Total 7 + 5 = 12.

To balance an equation, you need to find element that was not spread, in this problem it was Al and Na. You can split the equation into

? naoh →? na2so4

? al2 (so4) 3 + →? al (oh) 3

There is two Na on na2so4, so to balance it you need two NaOH

2 naoh → 1 na2so4

? al2 (so4) 3 + →? al (oh) 3

There is two Al in al2 (so4) 3 and one Al on al (oh) 3, so you need twice amount al (oh) 3 to balance the equation.

2 naoh → 1 na2so4

1 al2 (so4) 3 + → 2 al (oh) 3

In the equation above, there is 3 SO4 on reactant but 1 SO4 on the product, so multiply the naoh and na2so4 with 3

6 naoh → 3 na2so4

1 al2 (so4) 3 + → 2 al (oh) 3

The final equation would be:

1 al2 (so4) 3 + 6 naoh → 2 al (oh) 3 + 3 na2so4

The sum of the coefficient would be: 1+6+2+3=12