28 g of nitrogen and 6 g hydrogen were mixed in a 1 litre closed container. At equilibrim 17 g `NH_3` was produced. Calculate the weight of weight of nitrogen, hydrogen at equilibrium.

Explanation

Given `m_(N_2) = 28g, m_(H_2) = 6 g , V = 1L`
`(n_(N_2))_("initial")= 28/28 = 1mol`
`(n_(H_2))_("initial")= 6/2 = 3mol`
`N_2(g) + 3H_2(g) hArr 2NH_3(g)`

`[NH_3] = (17/17) = 1mol = 2x`
`rArr x = 0.5` mol
At equilibrium, `[N_2] = 1 - x = 0.5 mol`
`[H_2] = 3 -3x =3-3 (0.5) = 1.5 mol`
Weight of`N_2` = (no. of moles of `N_2`)` xx`molar mass of `N_2 = 0.5 xx 28 = 14g`
Weight of `H_2` = (no. of molesof `H_2`) `xx` molar mass of `H_2= 1.5 xx 2 = 3g`