Question

N2(g) + 3H2(g) →2NH3(g) If there is 10.02 g N2 and excess H2 present, the reaction yields 9.47 g NH3. Calculate the percent yield for the reaction

Answer 1

molar mass of N2 = 28 g/mol
number of mole of N2 = (given mass)/(molar mass)
= 10.02/28
= 0.358 mole
N2(g) + 3H2(g) →2NH3(g)
according to reaction
1 mole of N2 give 2 mole of NH3
0.358 mole of N2 give (2*0.358) mole of NH3
0.358 mole of N2 give 0.716 mole of NH3
number of mole of NH3 produce = 0.716 mole
mass of NH3 produce = (number of mole of NH3)*(molar mass of NH3)
= (0.716*17) g
= 12.2 g

%yield = {(actual yield)/(theoretical yield)}*100
= (9.47/12.2)*100
= 77.6 %

Answer : 77.6%