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?

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

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.