The total pressure of a gaseous mixture of `2.8 g N_(2)`, `3.2 g O_(2)`, and `0.5 g H_(2)` is `4.5 atm`. Calculate the partial pressure of each gas.

Explanation

Number of moles `=(Mass)/(Molar mass)`
Moles of `N_(2)(n_(N_(2)))=(2.8)/(28)=0.1mol`
Moles of `O_(2)(n_(O_(2)))=(3.2)/(32)=0.1 mol`
Moles of `H_(2)(n_(H_(2)))=(0.5)/(2)=0.25 mol`
`:. ` Total number of mole `=n_(N_(2))+n_(O_(2))+n_(H_(2))`
`=0.1+0.1+0.25=0.45`
Partial pressure a gas is
`(Number of moles)/(Total number of moles)xxTotal pressure`
`P_(N_(2))=(0.1)/(0.45)xx4.5atm=1.0 atm`
`P_(O_(2))=(0.1)/(0.45)xx4.5 atm=1.0 atm`
`P_(H_(2))=(0.25)/(0.45)xx4.5 atm=2.5 atm`