[miss.strw]

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!

[renge ishyo]

The balanced equation looks fine. The error I believe appears here:

68.12.00g/(32g/mol of O2) = 2.125 mol of O2

You do not have 68.12g of O2, you have 68.12g of NH3. So you need to divide 68.12g by the molecular weight of NH3. The approach in the calculation seems correct aside from this problem (just make sure the species cancel correctly and you should be fine).

[miss.strw]

thank you so much :]