Question

Question #461b0

Answer 1

`"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.