Question

Nitrogen and hydrogen combine at a high temperature, in the presence of a catalyst, to produce ammonia. N. (9)+3H,09 — 2 NH; (g) Assume 0.270 mol N, and 0.856 mol H, are present initially. After complete reaction, how many moles of ammonia are produced? NH, | How many moles of H, remain? H: How many moles of N, remain? moles of N, remain? N: What is the limiting reactant? O nitrogen hydrogen

Answer 1

1)
Balanced chemical equation is:
N2 + 3 H2 ---> 2 NH3 +


1 mol of N2 reacts with 3 mol of H2
for 0.27 mol of N2, 0.81 mol of H2 is required
But we have 0.856 mol of H2

so, N2 is limiting reagent
we will use N2 in further calculation

According to balanced equation
mol of NH3 formed = (2/1)* moles of N2
= (2/1)*0.27
= 0.54 mol
Answer: 0.540 mol

2)
According to balanced equation
mol of H2 reacted = (3/1)* moles of N2
= (3/1)*0.27
= 0.81 mol
mol of H2 remaining = mol initially present - mol reacted
mol of H2 remaining = 0.856 - 0.81
mol of H2 remaining = 4.6*10^-2 mol
Answer: 0.0460 mol

3)
Answer: 0.0 mol

4)
Answer: Nitrogen