Question

What mass of chlorine occupies 33.6 L at STP?

Answer 1

`"105 g"`

Explanation:

For starters, you should know that STP conditions are defined as a pressure of `"100 kPa"` and a temperature of `0^@"C"`.

Under these conditions, `1` mole of any ideal gas occupies `"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 `"22.72 L"`.

`"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, `"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 color(red)(cancel(color(black)("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.