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QUESTION

# When heated, KClO3 decomposes into KCl and O2. If this reaction produced 20.3 g of KCl, how much O2 was produced (in grams)?

"13.1 g O"_2

Start by writing the balanced chemical equation that describes the 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) that exists between the two products of the reaction to a gram ratio by using the molar masses of the two .

"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 , 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)|)))