Question

Nitropen and hydrogen combine at a high temperature, in the presence of a catalyst, to produce ammonia. N, (g) + 3H2(g) 2 NH,(Ⓡ) Assume 0.100 mol N, and 0.325 mol H, are present initially. After complete reaction, how many moles of ammonia are produced? How many moles of H, remain? How many moles of N, remain? What is the limiting reactant?

Answer 1

1)

Balanced chemical equation is:

N2 + 3 H2 ---> 2 NH3 +

1 mol of N2 reacts with 3 mol of H2

for 0.1 mol of N2, 0.3 mol of H2 is required

But we have 0.325 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)*0.1

= 0.2 mol

Answer: 0.200 mol

2)

According to balanced equation

mol of H2 reacted = (3/1)* moles of N2

= (3/1)*0.1

= 0.3 mol

mol of H2 remaining = mol initially present - mol reacted

mol of H2 remaining = 0.325 - 0.3

mol of H2 remaining = 2.5*10^-2 mol

Answer: 0.0250 mol

3)

N2 is limiting reagent

So, there will be no N2 remaining

Answer: 0.00 mol

4)

Answer: nitrogen