Balance the equation:
CuSO₄ + Na(OH) -----> Cu(OH)₂ + Na₂SO₄
We will start counting the amount of each element that we have on both sides of the equation. But, to make it easier, we will consider the OH⁻ ion as a whole, and the SO₄²⁻ as a whole. So:
__ CuSO₄ + ___ Na(OH) -----> ___ Cu(OH)₂ + ___ Na₂SO₄
Na: 1 Na: 2
OH: 1 OH: 2
Cu: 1 Cu: 1
SO₄: 1 SO₄: 1
Now, we will start changing the coefficients to balance the equation. The first one is Na, and we have 2 atoms on the right side and only one on the left. That tells us that we have to change the coefficient that is in front of NaOH, and write a 2 there. Let's do that:
__ CuSO₄ + 2 Na(OH) -----> ___ Cu(OH)₂ + ___ Na₂SO₄
Na: 2 Na: 2
OH: 2 OH: 2
Cu: 1 Cu: 1
SO₄: 1 SO₄: 1
After we change that coefficient we count again and now we have two atoms of Na on the left, so it is balanced. The OH wasn't balanced but when we changed that coefficient we balanced it. And Cu and SO4 were already balanced. The equation is balanced now.
The answer to our problem is:
CuSO₄ + 2 Na(OH) -----> Cu(OH)₂ + Na₂SO₄