# Question 75890

Jul 13, 2017

Here's what I got.

#### Explanation:

You can't really answer this question because you don't have any information about the pressure and the temperature at which the gas is collected.

If you assume that the gas is being collected at a temperature of $T$ $\text{K}$ and a pressure of $P$ $\text{atm}$, you can use the ideal gas law equation

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{P V = n R T}}}$

Here

• $P$ is the pressure of the gas
• $V$ is the volume it occupies
• $n$ is the number of moles of gas present in the sample
• $R$ is the universal gas constant, equal to $0.0821 \left(\text{atm L")/("mol K}\right)$
• $T$ is the absolute temperature of the gas

to calculate the number of moles of oxygen gas produced by the reaction.

Rearrange the ideal gas law equation to solve for $n$

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

and plug in your values to find

$n = \left(P \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{atm"))) * 11.2 color(red)(cancel(color(black)("L"))))/(0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * T color(red)(cancel(color(black)("K}}}}\right) = 136.42 \cdot \frac{P}{T}$ $\text{moles}$

Now, the balanced chemical equation that describes this decomposition reaction looks like this

$2 {\text{KClO"_ (3(s)) stackrel(color(white)(acolor(red)(Delta)aaa))(->) 2"KCl"_ ((s)) + 3"O}}_{2 \left(g\right)}$

Notice that for every $2$ moles of potassium chlorate that undergo decomposition you get $3$ moles of oxygen gas.

This means that in order for the reaction to produce the number of moles of oxygen gas present in your sample, it must consume

(136.42 * P/T) color(red)(cancel(color(black)("moles O"_2))) * "2 moles KClO"_3/(3color(red)(cancel(color(black)("moles O"_2)))) = (90.95 * P/T)# ${\text{moles KClO}}_{3}$

At this point, all you need to know are the conditions fo pressure and temperature at which the gas is being collected.

For example, at STP, which is defined as a pressure of $\frac{100}{101.325}$ $\text{atm}$ and a temperature of $\text{273.15 K}$, the number of moles of potassium chlorate will be equal to

${\text{moles of KClO"_3 = (90.95 * (100/101.325)/273.15) = "0.329 moles KClO}}_{3}$

The answer must be rounded to three sig figs, the number of sig figs you have for the volume of oxygen gas.