20.0 kg of `N_2(g)` and 3.0 kg of `H_2(g)` are mixed to produce `NH_3(g)` . The amount of `NH_3` (g) formed is

Options

A. 17 g
B. 34 g
C. 20 g
D. 3 kg

Answer

A

Explanation

`underset(28)(N_2) + underset(2 xx 3)(3H_2) to underset (34)(2NH_3)`
1 mole of `N_2` (28g) combine with 3 moles of `H_2(6g)`
` therefore ` 3kg of `H_2` can react with `28/6 xx 3 = 14 kg ` of `N_2 H_2` is limiting reagent .
Moles of `NH_3` formed ` = 34/6 xx 3 = 17 kg ` .