Question

One of the steps in the commercial process for converting ammonia to nitric acid is the conversion of NH3 to NO:

4NH3(g)+5O2(g)→4NO(g)+6H2O(g)

In a certain experiment, 1.20 g of NH3 reacts with 2.20 g of O2.

a) How many grams of NO and of H2O form? Enter your answers numerically separated by a comma.

b) How many grams of the excess reactant remain after the limiting reactant is completely consumed? Express your answer in grams to three significant figures.

Answer 1

a)

Molar mass of NH3,
MM = 1*MM(N) + 3*MM(H)
= 1*14.01 + 3*1.008
= 17.034 g/mol


mass(NH3)= 1.2 g

number of mol of NH3,
n = mass of NH3/molar mass of NH3
=(1.2 g)/(17.034 g/mol)
= 7.045*10^-2 mol

Molar mass of O2 = 32 g/mol


mass(O2)= 2.2 g

number of mol of O2,
n = mass of O2/molar mass of O2
=(2.2 g)/(32 g/mol)
= 6.875*10^-2 mol
Balanced chemical equation is:
4 NH3 + 5 O2 ---> 4 NO + 6 H2O


4 mol of NH3 reacts with 5 mol of O2
for 0.0704 mol of NH3, 0.0881 mol of O2 is required
But we have 0.0688 mol of O2

so, O2 is limiting reagent
we will use O2 in further calculation


Molar mass of NO,
MM = 1*MM(N) + 1*MM(O)
= 1*14.01 + 1*16.0
= 30.01 g/mol

According to balanced equation
mol of NO formed = (4/5)* moles of O2
= (4/5)*0.0688
= 5.5*10^-2 mol


mass of NO = number of mol * molar mass
= 5.5*10^-2*30.01
= 1.65 g
According to balanced equation
mol of H2O formed = (6/5)* moles of O2
= (6/5)*0.0688
= 8.25*10^-2 mol


mass of H2O = number of mol * molar mass
= 8.25*10^-2*18
= 1.49 g
Answer: 1.65, 1.49

b)
According to balanced equation
mol of NH3 reacted = (4/5)* moles of O2
= (4/5)*0.0688
= 5.5*10^-2 mol
mol of NH3 remaining = mol initially present - mol reacted
mol of NH3 remaining = 0.0704 - 0.055
mol of NH3 remaining = 0.0154 mol


Molar mass of NH3,
MM = 1*MM(N) + 3*MM(H)
= 1*14.01 + 3*1.008
= 17.034 g/mol

mass of NH3,
m = number of mol * molar mass
= 1.545*10^-2 mol * 17.034 g/mol
= 0.263 g
Answer: 0.263 g