Answer:
Ba^2+(aq) + 2OH-(aq) + 2H+(aq) + SO4^2-(aq) → BaSO4(s) + 2H2O(l)
Explanation:
Step 1: Data given
the contribution of protons from H2SO4 is near 100 %.
Step 2: The unbalanced equation
.Ba(OH)2(aq) + H2SO4(aq) → BaSO4(s) + H2O(l)
Step 3: Balancing the equation
On the left side we have 4x H (2x in Ba(OH)2 and 2x in H2SO4). On the right side, we have 2x H (in H2O).
To balance the amount of H on both sides, we have to muliply H2O (on the right side) by 2.
Ba(OH)2(aq) + H2SO4(aq) → BaSO4(s) + 2H2O(l)
Step 4: The net ionic equation
The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will.
Ba^2+(aq) + 2OH-(aq) + 2H+(aq) + SO4^2-(aq) → BaSO4(s) + 2H2O(l)
After canceling those spectator ions in both side, look like this:
Ba^2+(aq) + 2OH-(aq) + 2H+(aq) + SO4^2-(aq) → BaSO4(s) + 2H2O(l)
