Question

Question #7f200

Answer 1

`"16.0 g O"_2`

Explanation:

The thing to remember about STP conditions, i.e. Standard Temperature and Pressure, is that the volume occupied by exactly `1` mole of an ideal gas is equal to `"22.4 L"`.

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SIDE NOTE That value corresponds to the old definition of STP conditions, i.e. a pressure of `"1 atm"` and a temperature of `0^@"C"`.

In the current definition, the temperature is `0^@"C"`, but the pressure is `"100 kPa"`. Under these conditions, the volume occupied by `1` mole of an ideal gas is equal to `"22.7 L"`.

Sadly enough, not that many textbooks and online resources have adapted to the change, so the old value is still going strong. I'll use the old value here, but keep in mind that the correct value is not `"22.4 L"`.

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The volume occupied by exactly `1` mole of an ideal gas kept under STP conditions is called the molar volume of a gas at STP.

In your case, you know that `1` mole of oxygen gas occupies `"22.4 L"` at STP. You can use the molar volume of a gas at STP as a conversion factor to find the number of moles of oxygen gas present in your sample

`11.2 color(red)(cancel(color(black)("L O"_2))) * overbrace("1 mole O"_2/(22.4color(red)(cancel(color(black)("L O"_2)))))^(color(blue)("molar volume of a gas at STP")) = "0.500 moles O"_2`

To convert the number of moles to grams, use the molar mass of oxygen gas -- do not forget that oxygen gas exists as a diatomic molecule!

`0.500 color(red)(cancel(color(black)("moles O"_2))) * "32.0 g"/(1color(red)(cancel(color(black)("mole O"_2)))) = color(darkgreen)(ul(color(black)("16.0 g")))`

The answer is rounded to three sig figs.