Question

How many moles of which reactant will remain if 1.39 moles of N2 and 3.44 moles of
H2 will react to form ammonia? find out how many grams of ammonia can be formed
and how many moles of limiting reactant is required to completely exhaust the other
reactant that is in excess?

Answer 1

#74975 Chemistry, Other

How many moles of which reactant will remain if 1.39 moles of N₂ and 3.44 moles of H₂ will react to form ammonia? Find out how many grams of ammonia can be formed and how many moles of limiting reactant is required to completely exhaust the other reactant that is in excess?

Answer:

$$3 \mathrm{H}_2 + \mathrm{N}_2 = 2 \mathrm{NH}_3$$ $$n \left(\mathrm{N}_2\right) = \frac{1}{3} \cdot n \left(\mathrm{H}_2\right) = \frac{1}{3} \cdot 3.44 = 1.15 \mathrm{mol}$$

Therefore, H₂ will be a limiting reagent. It will react fully.

The amount of N₂ to be left is: n (N₂) = 1.39 - 1.15 = 0.24 mol

The amount of H₂ required to exhaust all the N₂ is: n (H₂) = (1.39 · 3) - 3.44 = 0.73 mol

The amount of ammonia to be formed from the existing amount of reactants is:

$$n \left(\mathrm{NH}_3\right) = \frac{2}{3} n \left(\mathrm{H}_2\right) = \frac{2}{3} \cdot 3.44 = 2.3 \mathrm{mol}$$ $$m = nM$$ $$M \left(\mathrm{NH}_3\right) = 18 \mathrm{g/mol}$$ $$m \left(\mathrm{NH}_3\right) = 2.3 \cdot 18 = 41.3 \mathrm{g}$$

Answer provided by AssignmentExpert.com