Question

If 5.12 g of oxygen O2 gas occupies a volume of 6.21L at a certain temperature and pressure, how many grams of oxygen gas will occupy 30.3 L under the same conditions ?

Answer 1

Answer : The mass of
`O_2` occupy 30.3 L under the same conditions will be, 24.9 grams.

Explanation :

First we have to calculate the moles of
`O_2`

`\text{Moles of }O_2=\frac{\text{Given mass }O_2}{\text{Molar mass }O_2}=(5.12g)/(32g/mol)=0.16mol`

Now we have to calculate the moles of
`O_2` in 30.3 L by using Avogadro's law.

Avogadro's law : It is defined as the volume of gas is directly proportional to the number of moles of gas at constant pressure and temperature.

`V\propto n`

or,

`(V_1)/(n_1)=(V_2)/(n_2)`

where,

`V_1` = initial volume of gas = 6.21 L

`V_2` = final volume of gas = 30.3 L

`n_1` = initial moles of gas = 0.16 mol

`n_2` = final temperature of gas = ?

Now put all the given values in the above equation, we get:

`(6.21L)/(0.16mol)=(30.3L)/(n_2)`

`n_2=0.781mol`

Now we have to calculate the mass of
`O_2`

`\text{ Mass of }O_2=\text{ Moles of }O_2* \text{ Molar mass of }O_2`

Molar mass of
`O_2` = 32 g/mol

`\text{ Mass of }O_2=(0.781moles)* (32g/mole)=24.9g`

Therefore, the mass of
`O_2` occupy 30.3 L under the same conditions will be, 24.9 grams.