Question #1935e

The balanced equation is:

`H_2SO_(4(aq))+2NH_4OH_((aq))rarr(NH_4)_2SO_4(aq)+2H_2O_((l))`

Sulfuric acid is diprotic so is able to give up 2 H+ ions. `NH_4OH` is not a compound that can actually exist as a discrete substance. It is better described as ammonia solution which is formed from dissolving ammonia gas in water:

`NH_(3(g))+ H_2O_((l))rightleftharpoonsNH_(4(aq))^(+)+OH_((aq))^-`

You are dealing with a reaction between a strong acid and a weak base, which results in the formation of salt and water. The reaction's general equation is

`H_2SO_(4(aq)) + 2NH_4OH_((aq)) -> (NH_4)_2SO_4(aq) + 2H_2O_((l))`

The complete ionic equation looks like this

`2H_((aq))^(+)+ SO_(4(aq))^(2-) + 2NH_(4(aq))^(+) + 2OH_((aq))^(-) -> 2NH_(4(aq))^(+) + SO_(4(aq))^(2-) + 2H_2O_((l))`

The net ionic equation you'll often see is

`H_((aq))^(+) + OH_((aq))^(-) -> H_2O_((l))`

However, sulfuric acid, which is a strong acid, reacts with ammonium hydroxide, which is a weak base, to produce salt, ammonium sulfate, and water.

Since it is a weak base, ammonium hydroxide will dissociate only slightly in water, so another version of the net ionic equation could be:

`H_((aq))^(+) + NH_4OH_((aq)) -> NH_(4(aq))^(+) + H_2O_((l))`

However, this does not change the balanced general reaction, which still stands as it was written in the beginning of the post.

`H_2SO_4+2NH_4OH->(NH_4)_2SO_4+2H_2O`

You see that at the right of the equation there are 2 `NH_4`-groups, while at the left there is only one.

To even this out you may only double the whole `NH_4OH`
which works out right, because then the 2 `H`'s from the `H_2SO_4` can combine with the 2 `OH`'s from the `NH_4OH`
to form 2 `H_2O`'s

In reality it all happens a bit differently, with ions and stuff like that, but for balancing the equation this is enough.