# Question 37cc4

Jun 18, 2017

Here's what I got.

#### Explanation:

You can't really find the volume of the oxygen gas because you're missing information on the conditions you have for pressure and temperature, but you can find the mass of oxygen gas produced by the reaction.

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

Notice that when $2$ moles of potassium chlorate undergo thermal decomposition, the reaction produces $3$ moles of oxygen gas.

Use the molar mass of the reactant to find the number of moles present in the $\text{245 g}$ sample

245 color(red)(cancel(color(black)("g"))) * "1 mole KClO"_3/(122.55color(red)(cancel(color(black)("g")))) = "1.9992 moles KCl"_3

So, you know that the number of moles of oxygen gas is given by the aforementioned $2 : 3$ mole ratio that exists between potassium chlorate and oxygen gas, so you can say that the reaction will produce

1.9992 color(red)(cancel(color(black)("moles KClO"_3))) * "3 moles O"_2/(2color(red)(cancel(color(black)("moles KClO"_3)))) = "2.9988 moles O"_2#

To convert this to grams, use the molar mass of oxygen gas

$2.9988 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles O"_2))) * "32.0 g"/(1color(red)(cancel(color(black)("mole O"_2)))) = color(darkgreen)(ul(color(black)("96.0 g}}}}$

The answer is rounded to three sig figs, the number fo sig figs you have for the mass of potassium chlorate.

Now, in order to calculate the volume occupied by the oxygen gas, you need to 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

Rearrange to solve for $V$

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

At this point, you would need the pressure and the temperature at which the gas is being collected.

For example, if you were to collect the gas at a pressure of $\text{1 atm}$ and a temperature of

${20}^{\circ} \text{C" = 20^@"C" = 273.15 = "293.15 K} \to$ room temperature

the volume occupied by the gas would be equal to

$V = \left(2.9988 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles"))) * 0.0821(color(red)(cancel(color(black)("atm"))) * "L")/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * 273.15color(red)(cancel(color(black)("K"))))/(1color(red)(cancel(color(black)("atm}}}}\right)$

$V = \text{72.2 L} \to$ three sig figs

Now, what would happen to the volume if you were to collect the gas at $\text{0.5 atm}$ and $\text{313 K}$ ?

$V = \left(2.9988 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles"))) * 0.0821(color(red)(cancel(color(black)("atm"))) * "L")/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * 313.15color(red)(cancel(color(black)("K"))))/(0.5color(red)(cancel(color(black)("atm}}}}\right)$

$V = \text{154 L} \to$ three sig figs

So as you can see, the actual value of the volume depends on the pressure and temperature at which the gas is being collected.