Question

What mass of `"potassium sulfate"` would result from the reaction between excess `"sulfuric acid"` and `"potassium hydroxide"`?

Answer 1

Not a good question...........as it is based on a faulty premise.

Explanation:

We need a stoichiometric equation to represent the reaction:

`H_2SO_4(aq) + 2KOH(aq) rarrK_2SO_4(aq) +2H_2O(l)`

And thus moles of potassium sulfate is equivalent to half the number of moles of potassium hydroxide.

`"Moles of potassium hydroxide"` `=` `(14*g)/(56.11*g*mol^-1)`

`=0.250*mol`

And by the stoichiometry of the rxn, we get `0.125*molxx174.26*g*mol^-1=??*g` potassium sulfate, and `2xx0.125*molxx18.01*g*mol^-1=??*g` water.

The number of molecules is simply the molar quantity multiplied by `N_A, "Avogadro's number", 6.022xx10^23*mol^-1`.

But in fact, all of these calculations are based on a FAULTY premise. The product of potassium hydroxide with EXCESS sulfuric acid is not `"potassium sulfate"`, `K_2SO_4`, but `"potassium bisulfate"`, i.e. `KHSO_4(aq)`. In the reaction as written we would get 1 equiv of `KHSO_4(aq)` and ONE of water:

`H_2SO_4(aq) + KOH(aq) rarr KHSO_4(aq) + H_2O(l)`.

So if you want to be bolshie, you can state that NO potassium sulfate results from the reaction.