What mass of chlorine occupies 33.6 L at STP?

Jun 15, 2017

$\text{105 g}$

Explanation:

For starters, you should know that STP conditions are defined as a pressure of $\text{100 kPa}$ and a temperature of ${0}^{\circ} \text{C}$.

Under these conditions, $1$ mole of any ideal gas occupies $\text{22.72 L}$. In other words, the molar volume of a gas at STP, i.e. the volume occupied by $1$ mole of gas, is equal to $\text{22.72 L}$.

$\text{1 mole ideal gas " stackrel(color(white)(acolor(red)("STP conditions")aaa))(color(blue)(->)) " 22.72 L}$

So, the first thing to do here is to figure out the number of moles of chlorine gas, ${\text{Cl}}_{2}$, present in your sample.

To do that, use the molar volume of a gas at STP

33.6 color(red)(cancel(color(black)("L"))) * "1 mole Cl"_2/(22.72color(red)(cancel(color(black)("L")))) = "1.479 moles Cl"_2

To convert this to grams, sue the molar mass of chlorine gas

$1.479 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles Cl"_2))) * "70.906 g"/(1color(red)(cancel(color(black)("mole Cl"_2)))) = color(darkgreen)(ul(color(black)("105 g}}}}$

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