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28 g of nitrogen and 6 g hydrogen were mixed in a1 litre closed container. At equilibrim 17 g NH_3 was produced. Calculate the weight of weight of nitrogen, hydrogen at equilibrium. |
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Answer» Solution :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` `rArrx = 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` |
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