Question

Ammonia, NH3 , reacts with oxygen to form nitrogen gas and water.

4NH3(aq)+3O2(g)⟶2N2(g)+6H2O(l)

If 3.15 g of NH3 reacts with 4.73 g of O2 and produces 0.750 L of N2 at 295 K and 1.00 atm , which reactant is limiting?

a. O2(g)

b. NH3(aq)

What is the percent yield of the reaction?

percent yield: %

Answer 1

4NH3(aq)+3O2(g)⟶2N2(g)+6H2O(l)

No. of mole of NH3 = (3.15 g) /(17.03 g/mol) = 0.185 mol

Moles of O2 = (4.73 g) / (32 g/mol) = 0.148 mol

No. of moles of oxygen needed to react with 0.185 mol of NH3 = (3 / 4) x (0.185 mol) = 0.138 mol

Here; oxygen present in excess.

So; the limiting reagent is NH3

Limiting reactant is NH3

No. of moles of product N2 formed = (2 / 4) x 0.185 mol = 0.0925 mol

theoretical weight of N2 produced = (0.0925 mol) x (28 g/mol) = 2.59 g

moles of N2 produced (n) = PV / RT = (1 atm x 0.75 L) / (0.0821 x 295 K) = 0.0309 mol

practical yield of N2 = (0.0309 mol) x (28 g/mol) = 0.867 g

Percent yield = (0.867 / 2.59) x 100 = 33.5 %