# Question 9fd5a

Jul 16, 2015

You'd need 138 g of potassium hydroxide to produce that many moles of potassium sulfate.

#### Explanation:

Potassium hydroxide, $K O H$, will react with sulfuric acid, ${H}_{2} S {O}_{4}$, to produce water and potassium sulfate, a soluble salt.

The balanced chemical equation for this neutralization reaction looks like this

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

Notice that you have a $\textcolor{red}{2} : 1$ mole ratio between potassium hydroxide and potassium sulfate. This means that very 2 moles of the former that react will produce 1 mole of the latter.

In other words, regardless of how many moles of potassium sulfate are produced, the reaction used up twice as many moles of potassium hydroxide.

Since you know how many moles of ${K}_{2} S {O}_{4}$ are produced, you can use the aforementioned mole ratio to figure out how many moles of $K O H$ reacted

1.23cancel("moles"K_2SO_4) * (color(red)(2)" moles "KOH)/(1cancel("mole"K_2SO_4)) = "2.46 moles" $K O H$

To get the mass needed to deliver this many moles to the reaction, use potassium hydroxide's molar mass

2.46cancel("mole") * "56.106 g"/(1cancel("mole")) = "138.02 g KOH"#

Rounded to three sig figs, the answer will be

${m}_{K O H} = \textcolor{g r e e n}{\text{138 g KOH}}$