# A sample of KClO_3 on decomposition gives 448 mL of oxygen gas at NTP. Calculate: (a) Weight of oxygen produced (b) Weight of KClO_3 originally taken (c) Weight of KCl produced?

Jul 18, 2015

#### Answer:

Mass of oxygen: 0.596 g
Mass of potassium chlorate: 1.52 g
Mass of potassium chloride: 0.927 g

#### Explanation:

Start with the balanced chemical equation for the decomposition of potassium chlorate, $K C l {O}_{3}$

$\textcolor{g r e e n}{2} K C l {O}_{3 \left(s\right)} \stackrel{\textcolor{red}{\text{heat}}}{\to} \textcolor{b l u e}{2} K C {l}_{\left(s\right)} + \textcolor{\mathmr{and} a n \ge}{3} {O}_{2 \left(g\right)}$

The important thing to notice here is that $\textcolor{g r e e n}{2}$ moles of potassium chlorate produce $\textcolor{b l u e}{2}$ moles of potassium chloride and $\textcolor{\mathmr{and} a n \ge}{3}$ moles of oxygen gas.

Since you know the volume and conditions under which the oxygen gas was produced, you can use the ideal gas law equation to determine how many moles were produced.

At NTP, or Normal Temperature and Pressure, the pressure is set at 1 atm and the temperature at ${20}^{\circ} \text{C}$.

$P V = n R T \implies n = \frac{P V}{R T}$

n_(O_2) = (1cancel("atm") * 448 * 10^(-3)cancel("L"))/(0.082(cancel("atm") * cancel("L"))/("mol" * cancel("K")) * (273.15 + 20)cancel("K")) = "0.01864 moles" ${O}_{2}$

Use oxygen's molar mass to determine how many grams were produced

$0.01864 \cancel{\text{moles") * "32.0 g"/(1cancel("mole")) = color(green)("0.596 g } {O}_{2}}$

SIDE NOTE Do not forget to convert the temperature to Kelvin and the volume to liters!

The mass of potassium chlorate that underwent decomposition can be found in a similar manner. First, determine the number of moles by using the mole ratio that exists between potassium chlorate and oxygen gas

0.01864cancel("moles"O_2) * (color(green)(2)" moles "KClO_3)/(color(orange)(3)" moles "O_2) = "0.01243 moles" $K C l {O}_{3}$

The mass of the compound before decomposition was

$0.01243 \cancel{\text{moles") * "122.55 g"/(1cancel("mole")) = color(green)("1.52 g } K C l {O}_{3}}$

The number of moles of potassium chlorate produced by the reaction will be equal to the number of moles of potassium chlorate

0.01243cancel("moles"KClO_3) * (color(blue)(2)" moles "KCl)/(color(green)(2)" moles "KClO_3) = "0.01243 moles" $K C l$

The mass of potassium chloride is

$0.01243 \cancel{\text{moles") * "74.55 g"/(1cancel("mole")) = color(green)("0.927 g } K C l}$

All three values are rounded to three sig figs, the number of sig figs you gave for the volume of oxygen gas produced.