1.

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

Answer»

17 g
34 g
20 g
3 kg

Solution :`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 `.


Discussion

No Comment Found

Related InterviewSolutions