Question

What is the correct net ionic equation, including all coefficients, charges, and phases, for the following set of reactants? Assume that the contribution of protons from H2SO4 is near 100 %.Ba(OH)2(aq)+H2SO4(aq)→Express your answer as a chemical equation.

Answer 1

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)