Question bc587

Aug 17, 2016

${\text{1.60 g O}}_{2}$

Explanation:

The most important thing to do here is to make sure that your chemical equation is balanced!

As you can see, the chemical equation given to you is not balanced, so start by writing out a balanced version.

$\textcolor{b l u e}{2} {\text{KNO"_ (3(s)) stackrel(color(white)(acolor(red)(Delta)aaa))(->) 2"KNO"_ (2(s)) + "O}}_{2 \left(g\right)}$

Now, the balanced chemical equation tells you that you have a $\textcolor{b l u e}{2} : 1$ mole ratio between potassium nitrate, ${\text{KNO}}_{3}$, and oxygen gas, ${\text{O}}_{2}$.

This tells you that for every $\textcolor{b l u e}{2}$ moles of potassium nitrate that undergo decomposition, the reaction produces $1$ mole of oxygen gas.

The problem provides you with a mass of potassium nitrate, so right from the start you know that you should convert this $\textcolor{b l u e}{2} : 1$ mole ratio to a gram ratio.

To do that, use the molar masses of potassium nitrate and of oxygen gas, respectively

M_("M KNO"_3) = "101.1 g mol"^(-1)

M_("M O"_2) = "32.0 g mol"^(-1)

This means that the aforementioned mole ratio is equivalent to

(color(blue)(2)color(red)(cancel(color(black)("moles KNO"_3))) * "101.1 g"/(1 color(red)(cancel(color(black)("mole KNO"_3)))))/(1 color(red)(cancel(color(black)("mole O"_2))) * "32.0 g"/(1color(red)(cancel(color(black)("mole O"_2))))) = 202.2/32.0 -> gram ratio

This tells you that for every $\text{202.2 g}$ of potassium nitrate that undergo decomposition, $\text{32.0 g}$ of oxygen gas are produced.

Your sample of potassium nitrate will thus produce

10.1 color(red)(cancel(color(black)("g KNO"_3))) * "32.0 g O"_2/(202.2color(red)(cancel(color(black)("g KNO"_3)))) = color(green)(|bar(ul(color(white)(a/a)color(black)("1.60 g O"_2)color(white)(a/a)|)))#

The answer is rounded to three sig figs.