if 264g of N2H4 and 106g of N2O4 are mixed, calculate the number of moles of the excess reactant
Molar mass of N2H4 = 32.0452 g/mol
264/32.0452= 8.24moles
Molar mass of N2O4= 92.011 g/mol
106/92.011=1.15moles
N2H4 + N2O4 → 3 N2 + 4 H2O
7(8.24)+7(1.15)
= 57.68+8.05=65.73moles
Comments
Leave a comment