Question

what is the theoretical yield of ammonia in grams if 17.15 grams of nitrogen gas and 10.95 grams of hydrogen gas are allowed to react?

Answer 1

molar mass of N2 = 28 g/mol
molar mass of H2 = 2 g/mol
number of mole = (given mass)/(molar mass)
number of mole of N2 = 17.15/28
= 0.6125 mole
number of mole of H2 = 10.95/2
= 5.475 mole

reaction taking place is
N2 + 3H2 --> 2NH3
according to reaction
1 mol of N2 required 3 mol of H2
0.6125 mol of N2 required 1.838 mol of H2
but we have 5.475 mol of H2
so, H2 is in excess
and N2 is limiting reagent

again,
1 mol of N2 give 2 mol of NH3
0.6125 mol of N2 give 1.225 mol of NH3
number of mol of NH3 formed = 1.225 mole

molar mass of NH3 = 17 g/mol
mass of NH3 formed = (number of mol of NH3 formed)*(molar mass of NH3)
= 1.225*17
= 20.825 g

Answer : 20.825 g