For the reaction
NH3 + O2 --> NO + H2O
how many grams of O2 are needed to completely react with 68.12 g of NH3?
ATTEMPT
First make hydrogens equal:
2NH3 + O2 --> NO + 3H2O
Then make nitrogens equal:
2NH3 + O2 --> 2NO + 3H2O
Then make oxygens equal:
2NH3 + (5/2)O2 --> 2NO + 3H2O
Then getting rid of fractions in eqn.:
(optional)
4NH3 + 5O2 --> 4NO + 6H2O
68.12.00g/(32g/mol of O2) = 2.125 mol of O2
2.125 mol of O2 * (4 mole of NH3/5mole of O2)
= 1.7 mole of NH3
So you need 1.7mol * 17g/mol = 28.9 g NH3
which is an incorrect answer, what did i do wrong? please help!thank you!