Question #461b0

1 Answer
Feb 11, 2018

#"0.31 g"#

Explanation:

The idea here is that under STP conditions, which are defined as a pressure of #"100 kPa"# and a temperature of #0^@"C"#, #1# mole of any ideal gas occupies #"22.71 dm"^3# #-># this is known as the molar volume of a gas at STP .

Since you know that

#"1 dm"^3 = 10^3 quad "cm"^3#

you can say that at STP, #1# mole of any ideal gas occupies #2.271 * 10^4# #"cm"^3#.

You can use the molar volume of a gas at STP to find the number of moles of oxygen as present in your sample.

#220 color(red)(cancel(color(black)("cm"^3))) * "1 mole O"_2/(2.271 * 10^4 color(red)(cancel(color(black)("cm"^3)))) = 9.687 * 10^(-3) quad "moles O"_2#

To convert the number of moles of oxygen gas to moles, you can use the molar mass of oxygen gas.

#9.687 * 10^(-3) color(red)(cancel(color(black)("moles O"_2))) * "32.0 g"/(1color(red)(cancel(color(black)("mole O"_2)))) = color(darkgreen)(ul(color(black)("0.31 g")))#

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