What mass of chlorine occupies 33.6 L at STP?

1 Answer
Jun 15, 2017

"105 g"105 g

Explanation:

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

Under these conditions, 11 mole of any ideal gas occupies "22.72 L"22.72 L. In other words, the molar volume of a gas at STP, i.e. the volume occupied by 11 mole of gas, is equal to "22.72 L"22.72 L.

"1 mole ideal gas " stackrel(color(white)(acolor(red)("STP conditions")aaa))(color(blue)(->)) " 22.72 L"1 mole ideal gas aSTP conditionsaaa−−−−−−−−−− 22.72 L

So, the first thing to do here is to figure out the number of moles of chlorine gas, "Cl"_2Cl2, 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.