# How many particles of KCIO_3 are needed to produce 8.14 grams of O_2?

## $K C l {O}_{3} \to K C l + {O}_{2}$

Apr 11, 2017

$1.02 \times {10}^{23}$ formula units (particles) are needed to produce $\text{8.14 g O"_2}$.

The process is described in the explanation.

#### Explanation:

Potassium chlorate, $\text{KClO"_3}$, is an ionic compound. The particles of an ionic compound are called formula units. There are $6.022 \times {10}^{23}$ formula units in one mole of an ionic substance.

${\text{2KClO}}_{3}$$\rightarrow$${\text{2KCl + 3O}}_{2}$

Determine the molar mass, $M$, of $\text{O"_2}$ .

$M {\text{O}}_{2} :$(2xx"15.999 g/mol O"_2)="31.998 g/mol O"_2"

Determine mol $\text{O"_2}$

Multiply given mass of $\text{O"_2}$ by its molar mass so that mol is in the numerator and mass is in the denominator, $\text{mol"/"grams}$.

8.14color(red)cancel(color(black)("g O"_2))xx(1"mol O"_2)/(31.998color(red)cancel(color(black)("g O"_2)))="0.25439 mol O"_2"

Determine mol $\text{KClO"_3}$

Multiply mol ${\text{O}}_{2}$ by the mol ratio from the balanced equation,

"2 mol KClO"_4/("3 mol O"_2") with $\text{O"_2}$ in the denominator so that it cancels.

0.25439color(red)cancel(color(black)("mol O"_2))xx(2"mol KClO"_3)/(3color(red)cancel(color(black)("mol O"_2)))="0.16959 mol KClO"_3"

Determine the number of formula units of $\text{KClO"_3}$.

Multiply mol $\text{KClO"_3}$ by $6.022 \times {10}^{23}$ formula units/mol.

0.16959color(red)cancel(color(black)("mol KClO"_3))xx(6.022xx10^23color(white)(.)"formula units KClO"_3)/(1color(red)cancel(color(black)("mol KClO"_3")))=1.02xx10^23$\text{formula units KClO"_3}$ (rounded to three significant figures)

Note: I kept some extra digits during the process to reduce rounding errors, but the final answer was rounded to three significant figures due to $\text{8.14 g O"_2}$.