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

May 11, 2017

#### Answer:

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

#### Explanation:

We need a stoichiometric equation to represent the reaction:

${H}_{2} S {O}_{4} \left(a q\right) + 2 K O H \left(a q\right) \rightarrow {K}_{2} S {O}_{4} \left(a q\right) + 2 {H}_{2} O \left(l\right)$

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

$\text{Moles of potassium hydroxide}$ $=$ $\frac{14 \cdot g}{56.11 \cdot g \cdot m o {l}^{-} 1}$

$= 0.250 \cdot m o l$

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 ${N}_{A} , \text{Avogadro's number} , 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$.

But in fact, all of these calculations are based on a FAULTY premise. The product of potassium hydroxide with EXCESS sulfuric acid is not $\text{potassium sulfate}$, ${K}_{2} S {O}_{4}$, but $\text{potassium bisulfate}$, i.e. $K H S {O}_{4} \left(a q\right)$. In the reaction as written we would get 1 equiv of $K H S {O}_{4} \left(a q\right)$ and ONE of water:

${H}_{2} S {O}_{4} \left(a q\right) + K O H \left(a q\right) \rightarrow K H S {O}_{4} \left(a q\right) + {H}_{2} O \left(l\right)$.

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