Question

1.30 g H2 is allowed to react with 10.4 g N2, producing 2.53 g NH3. What is the theoretical yield in grams for this reaction under the given conditions? What is the percent yield for this reaction under the given conditions?

Answer 1

number of mole = (given mass)/(molar mass)
molar mass of H2 = 2 g/mol
number of mole H2 = 1.3/2
= 0.65 mole
molar mass of N2 = 28 g/mol
number of mole N2 = 10.4/28
= 0.37 mole

reaction taking place is
N2 + 3H2 -- >2NH3   
according to reaction
1 mole of N2 required 3 mole of H2
0.37 mole of N2 required (3*0.37) mole of H2
0.37 mole of N2 required 1.11 mole of H2
but we have 0.65 mole of H2
hence, H2 is limiting reagent

now,
3 mole of H2 give 2 mole of NH3
1 mole of H2 give 2/3 mole of NH3
0.65 mole of H2 give (2/3)*0.65 mole of NH3
mole of NH3 produced = 0.43 mole
molar mass of NH3 = 17 g/mol
theoretical yield of NH3 = (number of mole of NH3)*(molar mass of NH3)
= (0.43*17) g
= 7.3 g

%yield of NH3 = {(actual yield)/(theoretical yield)}*100
= (2.53/7.3)*100
= 34.7%