Question

4NH3 (g)+5O2 (g)-->4NO (g)+6H2O (g)

What is the maximum mass of H2O that can be produced by combining 78.0 G of each reactant?

Answer 1

mass of NH3 = 78.0 g
molar mass of NH3 = 17 g/mol
mol of NH3 = (mass)/(molar mass)
= 78.0/17
= 4.579 mol

mass of O2 = 78.0 g
molar mass of O2 = 32.0 g/mol
mol of O2 = (mass)/(molar mass)
= 78.0/32.0
= 2.438 mol

Balanced chemical equation is:
4NH3 + 5O2 ---> 6H2O + 4NO

4 mol of NH3 reacts with 5 mol of O2
for 4.579 mol of NH3 9.158 mol of O2 is required
But we have 2.438 mol of O2

O2 is limiting reagent
we will use O2 in further calculation

According to balanced equation
mol of H2O formed = (6/5)* moles of O2
= (6/5)*2.438
= 2.925 mol

mass of H2O = number of mol * molar mass
= 2.925*18.000
= 52.650 g

Answer: 52.7 g