KI+H2SO4+KIO3
Write the balanced net ionic equation.
5I^- + IO3^- + 6H^+ ==> 3I2 + 3H2O
You can add the phases if that is needed.
To write the balanced net ionic equation for the reaction between potassium iodide (KI), sulfuric acid (H2SO4), and potassium iodate (KIO3), we will start by writing the overall balanced chemical equation:
2 KI + H2SO4 + KIO3 -> K2SO4 + H2O + 2 I2
Now, we will separate the soluble ionic substances into their respective ions:
2 K+(aq) + 2 I-(aq) + 2 H+(aq) + SO4^2-(aq) + K+(aq) + IO3^-(aq) -> K2SO4(aq) + H2O(l) + 2 I2(aq)
Finally, we can cancel out the spectator ions (ions that appear on both sides of the equation without undergoing a change) to write the net ionic equation:
2 I-(aq) + 2 H+(aq) + IO3^-(aq) -> H2O(l) + 2 I2(aq)
Therefore, the balanced net ionic equation for the reaction is:
2 I-(aq) + 2 H+(aq) + IO3^-(aq) -> H2O(l) + 2 I2(aq)
To write the balanced net ionic equation, we first need to know the formulas of the compounds involved. The symbols "KI," "H2SO4," and "KIO3" represent the chemicals potassium iodide, sulfuric acid, and potassium iodate, respectively.
The balanced chemical equation for the reaction can be written as follows:
2 KI + H2SO4 + KIO3 ⟶ I2 + H2O + K2SO4
To write the net ionic equation, we remove the spectator ions, which are the ions that exist in the same form on both sides of the equation. In this case, the spectator ions are K+ and SO4^2-. Therefore, they can be removed from the equation:
2 I- (aq) + H+ (aq) + IO3- (aq) ⟶ I2 (s) + H2O (l)
This is the balanced net ionic equation for the reaction between potassium iodide, sulfuric acid, and potassium iodate.