Question

Question #32c09

Answer 1

`"1.37 L"`

Explanation:

Your starting point here will be the balanced chemical equation that describes this decomposition reaction

`color(blue)(2)"KClO"_ (3(s)) stackrel(color(white)(acolor(red)(Delta)aaa))(->) 2"KCl"_ ((s)) + color(purple)(3)"O"_ (2(g))` `uarr`

Notice that you have a `color(blue)(2):color(purple)(3)` mole ratio between potassium chlorate and oxygen gas.

This tells you that the reaction produces `3/2` times more moles of oxygen gas than moles of potassium chlorate consumed by the reaction.

Use potassium chlorate's molar mass to determine how many moles you have in that `"5.00-g"` sample

`5.00 color(red)(cancel(color(black)("g"))) * "1 mole KClO"_3/(122.55color(red)(cancel(color(black)("g")))) = "0.04080 moles KClO"_3`

Use the aforementioned mole ratio to determine how many moles of oxygen gas would be produced by this many moles of potassium chlorate

`0.04080color(red)(cancel(color(black)("moles KClO"_3))) * (color(purple)(3)color(white)(a)"moles O"_2)/(color(blue)(2)color(red)(cancel(color(black)("moles KClO"_3)))) = "0.0612 moles O"_2`

Now, STP conditions are usually defined as a pressure of `"1 atm"` and a temperature of `0^@"C"`. Under these conditions for pressure and temperature, one mole of any ideal gas occupies `"22.4 L"` `->` this is known as the molar volume of a gas at STP.

Use the molar volume as a conversion factor to determine how many liters would correspond to that many moles of oxygen gas

`0.0612color(red)(cancel(color(black)("moles O"_2))) * "22.4 L"/(1color(red)(cancel(color(black)("mole O"_2)))) = color(green)(|bar(ul(color(white)(a/a)"1.37 L"color(white)(a/a)|)))`

The answer is rounded to three sig figs, the number of sig figs you have for the mass of potassium chlorate.

SIDE NOTE It's worth mentioning that the definition of STP has changed to a pressure of `"100 kPa"` and a temperature of `0^@"C"`.

Under these conditions, one mole of any ideal gas occupies `"22.7 L"`.

However, since most textbooks and online resources still use the old definition, I used `"22.4 L"` in my calculations.