What mass of chlorine occupies 33.6 L at STP?

1 Answer
Jun 15, 2017

Answer:

#"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.