28 g of nitrogen and 6 g of hydrogen were mixed in a 1 litre closed container. At equilibrium 17 g `NH_(3)` was produced. Calculate the weight of nitrogen, hydrogen at equilibrium.

Explanation

Given `m_(N_(2))=28g,m_(H_(2))=6g,`
`V=1L`
`(n_(N_(2)))_("Initial")=(28)/(28)="1 mol"`
`(n_(H_(2)))_("Initial")=6/2=3" mol"`
`N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g)`

`[NH_(3)]=((17)/(17))="1 mol"`
Weight of `N_(2)` = (no. of moles of `N_(2)` ) `xx` molar mass of `N_(2)`
`=0.5xx28=14g`
Weight of `H_(2)` = (no. of moles of `H_(2)` ) `xx` molar mass of `H_(2)`
`=1.5xx2=3g`