# Question c75aa

Jul 25, 2015

mass of potassium hydroxide ($K O H$)
= (1.23 x 2) (39 + 1 + 16) $g$
= 137.76 $g$

#### Explanation:

$2 K O H + {H}_{2} S {O}_{4} \to {K}_{2} S {O}_{4} + 2 {H}_{2} O$

Jul 25, 2015

Mass of potassium hydroxide needed: 138 g

#### Explanation:

Start by writing the balanced chemical equation for this neutralization reaction

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

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

Simply put, for a number of moles of potassium sulfate produced, twice as many moles of potassium hydroxide reacted.

You know that the reaction produced 1.23 g of potassium sulfate, which means that the number of moles of potassium hydroxide that reacted is

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 of potassium hydroxide that would contain this many moles, use the compound's molar mass.

A substance's molar mass tells you what the mass of 1 mole of a substance is.

$2.46 \cancel{\text{moles"KOH) * "56.11 g"/(1cancel("mole"KOH)) = color(green)("138 g } K O H}$