# Question eda5e

Jul 14, 2015

#### Explanation:

The idea behind this problem is that you need to use the mass of the potassium perchlorate precipitate in order to figure out how many moles of potassium oxide were present in the sample.

The first thing you need to do is determine what the mass of the dry sample is. To do that, use the fact that it contains 20.5% water, which means that you get 20.5 g of water for every 100 g of sample.

0.765cancel("g sample") * "20.5 g water"/(100cancel("g sample")) = "0.1568 g water"

This means that the dry sample has a mass of

${m}_{\text{dry" = 0.765 - 0.1568 = "0.6082 g}}$

You know the mass of the precipitate, which means that you can use its molar mass to determine how many moles were formed by the reaction.

0.3822cancel("g") * ("1 mole "KClO_4)/(138.55cancel("g")) = "0.002759 moles" $K C l {O}_{4}$

Here is what I think takes place as far as the chemical reactions that form the precipitate are concerned. Potassium oxide, ${K}_{2} O$, can be dissolved in water to form potassium hydroxide, $K O H$.

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

To get the potassium perchlorate precipitate, you'd then react this solution with perchloric acid, $H C l {O}_{4}$.

$K O {H}_{\left(a q\right)} + H C l {O}_{4 \left(a q\right)} \to K C l {O}_{4 \left(s\right)} \downarrow + {H}_{2} {O}_{\left(l\right)}$

Notice that you have a $1 : 1$ mole ratio between potassium perchlorate and potassium hydroxide. This tells you that, in order for the reaction to produce 0.002759 moles of the former, an equal number of moles of the latter must react.

0.002759cancel("moles"KClO_4) * "1 mole KOH"/(1cancel("mole"KClO_4)) = "0.002759 moles" $K O H$

Now look at the first chemical equation. This time you have a $1 : \textcolor{red}{2}$ mole ratio between potassium oxide and potassium hydroxide.

This means that, in order to produce that many moles of potassium hydroxide, the sample must contain half as many moles of potassium oxide.

0.002759cancel("moles"KOH) * ("1 mole "K_2O)/(color(red)(2)cancel("moles"KOH)) = "0.001380 moles" ${K}_{2} O$

Now use potassium oxide's molar mass to figure out what mass contained this many moles

0.001380cancel("moles") * "94.196 g"/(1cancel("mole")) = "0.1300 g"

The percent composition of potassium oxide in the dry sample will thus be

(0.1300cancel("g"))/(0.6082cancel("g")) * 100 = 21.37% = color(green)("21.4%")#

I think maybe some rounding I did along the way caused my answer to be a little off, but it's as close as it gets.