Question

Question #dd20a

Answer 1

`"24.5 g"`

Explanation:

The first thing to do here is to write a balanced chemical equation that describes this decomposition reaction

`color(red)(2)"KClO"_ (3(s)) -> 2"KCl"_ ((s)) + color(blue)(3)"O"_ (2(g)) uarr`

Notice that it takes `color(red)(2)` moles of potassium chlorate to produce `color(blue)(3)` moles of oxygen gas.

This means that if you know how many moles of oxygen gas were produced by the reaction, you can backtrack and use this `color(red)(2):color(blue)(3)` mole ratio to figure out how many moles of potassium chlorate underwent decomposition.

Now, STP conditions, which were probably given to you as a pressure of `"1 atm"` and a temperature of `0^@"C"`, are characterized by the fact that one mole of any ideal gas occupies `"22.4 dm"^3` -- this is known as the molar volume of a gas at STP.

Use this value to calculate how many moles of oxygen gas were produced by the reaction

`6.72 color(red)(cancel(color(black)("dm"^3))) * " mole O"_2/(22.4color(red)(cancel(color(black)("dm"^3)))) = "0.30 moles O"_2`

This means that the decomposition reaction consumed

`0.30 color(red)(cancel(color(black)("moles O"_2))) * (color(red)(2)color(white)(a)"moles KClO"_3)/(color(blue)(3)color(red)(cancel(color(black)("moles O"_2)))) = "0.20 moles KClO"_3`

Now all you have to do is use the molar mass of potassium chlorate to calculate how many grams would contain this many moles

`0.20 color(red)(cancel(color(black)("moles KClO"_3))) * "122.5 g"/(1color(red)(cancel(color(black)("mole KClO"_3)))) = color(green)(|bar(ul(color(white)(a/a)color(black)("24.5 g")color(white)(a/a)|)))`

The answer is rounded to three sig figs.