Question

A gas sample at stp contains 1.15 g oxygen gas and 1.55 g nitrogen gas.what is the volume of the gas sample?

Answer 1

Step-by-step explanation:

The given data is as follows.

Mass of
`O_(2)` = 1.15 g

Mass of
`N_(2)` = 1.55 g

Therefore, moles of oxygen present will be as follows.

No. of moles of
`O_(2)` =
`\frac{mass}{\text{molar mass}}`

=
`(1.15 g)/(32 g/mol)`

= 0.035 mol

No. of moles of
`N_(2)` =
`\frac{mass}{\text{molar mass}}`

=
`(1.55 g)/(28.02 g/mol)`

= 0.055 mol

Hence, total no. of moles = moles of
`O_(2)` + moles of
`N_(2)`

= (0.035 + 0.055) mol

= 0.09 mol

Now, it is known that at STP volume is 22.4 L/mol. Hence, volume of the gas sample at STP for 0.09 moles will be as follows.

`0.09 mol * 22.4 L/mol`

= 2.01 L

Thus, we can conclude that volume of the given gas sample is 2.01 L.