# Question 09a86

Jul 19, 2017

(B) $2$ ${\text{mol KClO}}_{3}$

#### Explanation:

We're asked to find the number of moles of ${\text{KClO}}_{3}$ that need to decompose to yield $67.2$ ${\text{L O}}_{2}$ (assuming 100% yield).

The decomposition reaction of potassium chlorate is

$2 {\text{KClO"_3(s) rarr 2"KCl"(s) + 3"O}}_{2} \left(g\right)$

At s.t.p., one mole of any (ideal) gas occupied a volume of $22.41$ $\text{L}$, so we can use this to convert from liters of oxygen to moles of oxygen:

67.2cancel("L O"_2)((1color(white)(l)"mol O"_2)/(22.41cancel("L O"_2))) = color(red)(3.00 color(red)("mol O"_3

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

color(red)(3.00)cancel(color(red)("mol O"_2))((2color(white)(l)"mol KClO"_3)/(3cancel("mol O"_2))) = color(blue)(2 color(blue)("mol KClO"_3#

Thus, option (B) is correct.