Question

Question #cfef7

Answer 1

At STP, 1 mole of any ideal gas occupies exactly `"22.4 L"` - this is known as the molar volume of a gas at STP. You were given `"15.0 L"`, which automatically means that you have less than 1 mole.

`"15.0 L" * ("1 mole")/("22.4 L") = "0.670 moles oxygen"`

The link between moles and grams (and vice versa) is made through molar mass; a compound's (or an element's) molar mass expresses the weight in grams of 1 mole of that particular compound.

Since oxygen gas exists as a diatomic molecule - `"O"_2`, its molar mass will be twice that of one oxygen atom, which has a molar mass of `"16.0 g/mol"`.

Therefore, the mass of oxygen in grams will be

`"0.670 moles O"_2 * ("32.0 "g)/("1 mole O"_2) = "21.4 g"`

Answer 2

The mass of oxygen gas is 21.43g

At STP 1 mole of gas has a volume of 22.4 L

Oxygen has the molecular formula `O_2`.

The `A_r` of oxygen = 16

So the `M_r` of `O_2=(2xx16)=32`

So 1 mole of `O_2` weighs 32g

This means that 22.4 L weighs 32g

So 1L weighs `(32)/(22.4)g`

So 15 L weighs `(32)/(22.4)xx15=21.43g`