Question

Assume that you have 1.39 mol of N2 and 3.44 mol of H2.
i) How many moles of ammonia can you make?
ii) How many grams of which reactant will be left over.

Answer 1

Answer:

N2 + 3H2----2NH3

H2 is limiting reagent

3molH2-----2molNH3

1mol H2-----2/3molNH3

3.44mol H2----2×3.44/3=2.29mol

amount =2.29×17=39gm

N2 left 1.39-1.14=0.24mol

0.24×28=almost 6gm