What is the net ionic equation for the reaction that occures between nitrouse acid and strontium hydroxide ?
2HNO2 + Sr(OH)2 ==> Sr(NO2)2 + 2H2O
net ionic is
2H^+ + 2OH^- ==> 2H2O
To find the net ionic equation for the reaction between nitrous acid (HNO2) and strontium hydroxide (Sr(OH)2), we first need to write the balanced chemical equation.
The balanced chemical equation for the reaction is:
2HNO2 + Sr(OH)2 -> Sr(NO2)2 + 2H2O
Now, let's write the complete ionic equation by separating all the soluble ionic compounds into their respective ions:
2H+(aq) + 2NO2-(aq) + Sr^2+(aq) + 2OH-(aq) -> Sr^2+(aq) + 2NO2-(aq) + 2H2O(l)
Next, we can remove the spectator ions, which are the ions that appear on both sides of the equation without undergoing any changes. In this case, the Sr^2+ and NO2- ions are spectator ions. Removing them, we get the net ionic equation:
2H+(aq) + 2OH-(aq) -> 2H2O(l)
So, the net ionic equation for the reaction between nitrous acid and strontium hydroxide is 2H+(aq) + 2OH-(aq) -> 2H2O(l).
To determine the net ionic equation for the reaction between nitrous acid (HNO2) and strontium hydroxide (Sr(OH)2), we need to first write out the balanced chemical equation, and then identify the spectator ions that do not participate in the reaction.
The balanced chemical equation for the reaction can be written as:
2 HNO2 + Sr(OH)2 → Sr(NO2)2 + 2 H2O
To derive the net ionic equation, we need to split the reactants and products into their respective ions (if they are soluble) and cancel out the spectator ions. Here's the breakdown:
Reactants:
2 H+ + 2 NO2- + Sr2+ + 2 OH-
Products:
Sr2+ + 2 NO2- + 2 H2O
From this breakdown, we can see that the ions NO2- are present on both sides of the equation, making them spectator ions. So, we can remove them from the equation.
Thus, the net ionic equation for the reaction between nitrous acid and strontium hydroxide is:
2 H+ + Sr2+ → Sr(NO2)2 + 2 H2O