What volume of nitrogen at S.T.R can be obtained from a mixture of 10 g each of NH_4CI and NaNO_2.

What volume of nitrogen at S.T.R can be obtained from a mixture of 10

1.	What volume of nitrogen at S.T.R can be obtained from a mixture of 10 g each of NH_4CI and NaNO_2.
Answer» Solution :The CHEMICAL equations involved are: `underset(53.492 g)(NH_(4)Cl) + underset(69 g)(NaNO_(2)) to underset(64.052 g)(NH_(4)NO_(2)) + NaCl` `underset(64.052 g)(NH_(4)NO_(2)) to underset(22.4 "L at S.T.P.")(N_(2)) + 2H_(2)O` According to these equations, 53.492 g of `NH_4CI` react with 69 g of `NaNO_2` to FORM 22.4 L of `N_2` at S.T.R Since, given mixture contains 10 g each of `NH_4CI` and `NaNO_2`, it is obvious that `NH_4CI` is in excess and some PART of it will be left behind. Hence, `NaNO_2` is the limiting reagent in this case. `therefore 69 g` of `NaNO_(2)` from `N_(2) = 22.4 L` at S.T.P `therefore 10 g` of `NaNO_(2)` will form `N_(2) = (22.4)/69 xx 10 = 3.25 L` THUS, the given mixture will GIVE 3.25 L at S.T.P.