# How many grams of potassium chlorate decompose to potassium chloride and 797 mL of O2 at 128 degrees Celsius and 747 torr? 2KClO3 (s) = 2KCl (s) + 3O2 (g)

Jul 8, 2017

$1.95$ ${\text{g KClO}}_{3}$

#### Explanation:

We're asked to find the mass, in $\text{g}$, of ${\text{KClO}}_{3}$ that decomposes to give a certain amount of ${\text{O}}_{2}$.

Let's use the ideal gas equation to find the moles of ${\text{O}}_{2}$ that form.

P = 747cancel("torr")((1color(white)(l)"atm")/(760cancel("torr"))) = 0.983 $\text{atm}$

V= 797cancel("mL")((1color(white)(l)"L")/(10^3cancel("mL"))) = 0.797 $\text{L}$

$T = {128}^{\text{o""C}} + 273 = 401$ $\text{K}$

Plugging in known values, we have

$n = \frac{P V}{R T} = \left(\left(0.983 \cancel{\text{atm"))(0.797cancel("L")))/((0.082057(cancel("L")•cancel("atm"))/("mol"•cancel("K")))(401cancel("K}}\right)\right) = 0.0238$ ${\text{mol O}}_{2}$

Now, we'll use the coefficients of the chemical equation to find the relative number of moles of ${\text{KClO}}_{3}$ that reacted:

0.0238cancel("mol O"_2)((2color(white)(l)"mol KClO"_3)/(3cancel("mol O"_2))) = 0.0159 ${\text{mol KClO}}_{3}$

Finally, we'll use the molar mass of potassium chlorate ($122.55$ $\text{g/mol}$) to find the number of grams that reacted:

0.0159cancel("mol KClO"_3)((122.55color(white)(l)"g KClO"_3)/(1cancel("mol O"_2))) = color(blue)(1.95 color(blue)("g KClO"_3