1) Chemical reaction: 3H2 + N2 -> 2NH3
2) proportions 3 moles H2 ; 1 mol N2 : 2 mol NH3
3) molar masses
H2: 2*1g/mol = 2 g/mol
N2: 2* 14 g/mol = 28 g/mol
NH3: 14g/mol + 3 g/mol = 17 g/mol
4) Conversion of the data in grams to mol
#mol = mass / molar mass
# mol H2 = 15 g / 2 g/mol = 7.5 mol
# mol N2 = 25 g / 28 g/mol = 0.89 mol
5) Compare the ratio of the reagents with the theoretical ratio:
Reagents: 0.89 mol N2 : 7.5 mol H2
Theoretical: 1mol N2 : 3 mol H2
You conclude that the H2 is in excess and need to calculate how many moles react with 0.89 mol of N2.
6) moles of N2 that react:
x mol H2 / 0.89 mol N2 = 3 mol H2 / 1 mol N2 => x = 0.89*3/1 = 2.67 mol H2
7) Calculate the excess as Initial # mol H2 - # mol that reacted = 7.5 mol - 2.67 mol = 4.83 mol H2 in excess
8) Convert mol of H2 to grams of H2
mass = # mol * molar mass
mass = 4.83 mol * 2 g/mol = 9.66 grams in excess
