Question

A mixture of methane and oxygen gases at a total pressure of 727 mm Hg contains methane at a partial pressure of 571 mm Hg. If the gas mixture contains 3.62 grams of methane, how many grams of oxygen are present

Answer 1

Answer: 1.95 g of oxygen is present.

Step-by-step explanation:

According to Dalton's law, the total pressure is the sum of individual pressures.

`p_(total)=p_(methane)+p_(oxygen)`

Given :
`p_(total)` =total pressure of gases = 727 mm Hg

`p_(methane)` = partial pressure of methane = 571 mm Hg

`p_(oxygen)` = partial pressure of oxygen = ?

`727=571+p_(oxygen)`

`p_(oxygen)=156mmHg`

Also
`p_(oxygen)=x_(oxygen)* p_(total)`

Given : 3.62 g of methane is present

moles of methane =
`\frac{\text{Given mass}}{\text {Molar mass}}=(3.62g)/(16g/mol)=0.226moles`

`x_(oxygen)` = mole fraction of oxygen

=
`\frac{\text {moles of oxygen}}{\text {total moles}}=(y)/(y+0.226)`

`156=(y)/(y+0.226)* 727`

`y=0.061`

mass of oxygen =
`moles* {\text {Molar mass}}=0.061* 32=1.95g`

Thus 1.95 g of oxygen is present.