Question

Question #186eb

Answer 1

`"13.1 g O"_2`

Explanation:

Start by writing the balanced chemical equation that describes the decomposition of potassium chlorate, `"KClO"_3`, to form potassium chloride, `"KCl"`, and oxygen gas, `"O"_2`

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

Notice that for every two moles of potassium chlorate that undergo decomposition, you get `color(red)(2)` moles of potassium chloride and `color(blue)(3)` moles of oxygen gas.

You can convert this `color(red)(2):color(blue)(3)` mole ratio that exists between the two products of the reaction to a gram ratio by using the molar masses of the two compounds.

`"For KCl": " " M_M = "74.55 g mol"^(-1)`

`"For O"_2: " " M_M = "32.0 g mol"^(-1)`

This tells you that every mole of potassium chloride has a mass of `"74.55 g"`, which implies that `color(red)(2)` moles will have a mass of

`2 color(red)(cancel(color(black)("moles KCl"))) * "74.55 g"/(1color(red)(cancel(color(black)("mole KCl")))) = "149.1 g"`

For oxygen gas, every mole has a mass of `"32.0 g"`, which means that `color(blue)(3)` moles will have a mass of

`3 color(red)(cancel(color(black)("moles O"_2))) * "32.0 g"/(1color(red)(cancel(color(black)("mole O"_2)))) = "96.0 g"`

The `color(red)(2):color(blue)(3)` mole ratio will be equivalent to a `149.1 : 96.0` gram ratio, i.e. you get `"96.0 g"` of oxygen gas for every `"149.1 g"` of potassium chloride produced by the reaction.

This means that `"20.3 g"` of potassium chloride would correspond to

`20.3color(red)(cancel(color(black)("g KCl"))) * "96.0 g O"_2/(149.1color(red)(cancel(color(black)("g KCl")))) = "13.07 g O"_2`

Rounded to three sig figs, the number of sig figs you have for the mass of potassium chloride, the answer will be

`m_(O_2) = color(green)(|bar(ul(color(white)(a/a)"13.1 g"color(white)(a/a)|)))`