In order to find the limiting and/ or excess reactant, you have to first convert grams to moles, then once you have that you did the #s by the smallest mole #. Once you have that, you compare that to the coefficients of the balanced equation.
This was my answer, but it was wrong:
2NO + O2 ----> 2NO2
Molar Mass NO2 = 46g/mol
Molar mass NO = 30g/mol
Molar Mass O2 = 32g/mol
0.333 mols NO x 46g NO2/1mol NO = 15.3g NO2
mols NO = 10g NO/30g/1mol NO = 0.333 mols NO
mols O2 = 20g O2/32g O2/1mol O2 = 0.625 mols O2
Where did I mess up?? And how do I figure out what is limiting reactant and what is excess?
Am i solving this correctly, according to my answer/posts above??
Ok, so you just do .333/.333 =1 (NO)
and .625/.333= 1.87 (O2)
So, then you go back to your equation, and you see how 1 < 2(# of moles of NO in the equation), that means it is the limiting reactant
Then you do the same thing with 1.87, and 1.87> 1( # of mole of O2), so that tells you it is the excess reactant. Make sense?
I guess so, thanks "K"
Tmofey, you didn't do anything wrong in your chemistry. I think the problem is math. I see 20g and 30g as having 2 significant figures so you can't report more than 2 in your answer. If you are keying this into a database, as I suspect, those things are notoriously unforgiving for the number of s.f. So your answer is 15.3 g NO2, you are allowed only 2 s.f., round the answer to 15 g NO2 produced. I'll bet the database accepts that answer.
I will do that, thanks for your help once again DrBob222.