ok, so I got a test back today in my chemistery class, and I just barely got a d (69%). 10% of that came bacause I couldn't remember the life story of Ernist Ruthiford and every experiment he ever performed. this has _nothing_ to do with an understanding of chemistery but I digress... the thing that's realy got me unable to sleep now is this. there was a 5 point question pertaining to part of a limiting reactant problem, I honestly **** up the first part, I wrote down aluminum oxide as AlO, when it should have been Al2O3, so starting off with the wrong (but balenced) formula of
2AL + O2 -> 2AlO
and told that I had a mass of aluminum and oxygen that I can't remember (I think it was something like 80g Al and 117g O2, that isn't the right number but we'll go with it because it proves the point) I need to find the limiting reactant. the way the teacher want's it done is thus:
convert mass to moles of each reactant then find the potential moles of one of the products useing those moles assumeing everything else is in exsess, for each reactant, then compare the moles of that particular product between all the reactants and find the one that is smallest and that is the limiting reactant.
what I did was:
I find the number of moles of each reactant, treating the reactant as a whole unit, that is I treat 2Al as two alluminums, I get the moles of each of these and compare them, in essence I assume one 'product' is made and the ratios tell me how much of each I need to make it.
her math would go as such:
80(the mass of the aluminum)/27(aproximation of the atomic mass of aluminum)=2.96(moles of aluminum)
2.96*1(for every Al atom I get 1 AlO)=2.96(moles of AlO)
117(mass of the oxygen)/32(molecular mass of O2)=3.65(moles of O2)
3.65*2(for every O2 molucule I get 2 AlO)=7.31(moles of AlO)
2.96(moles from AL) < 7.31(moles from O2) -> Al is the limiting reactant
my math is as follows:
80(mass of the aluminum)/54(mass of two aluminum attoms)=1.48(moles of 'product' (product in this case is 2AlO))
117(mass of the oxygen)/32(molecular mass of one oxygen molocule)=3.65(moles of 'product' (2AlO))
1.48(from aluminum)<3.65(from oxygen) -> Al is the limiting reactant
I came to the same answer with a diferent, but logicaly sound reasoning, the only thing I did diferent is I factored out the ****ing two as to not do twice as much unnessisary math. in her defence I didn't write out a product of 2Al anywere and as soon as I found the mass of aluminum and started to do the division I realised it was already smaller than oxygen so I just wrote out'/2 would be even smaller' and didn't actualy write out 1.48. I spend nearly all hour trying to explain it to her, and I get the destinct impression she didn't beleive me and thought I was trying to BS my way out of it. now she says 'I'll ask some of the other teachers there oppionions and see what they say' and I'm ever-so-sure she's going to actualy explain my point properly and not simply tell them I did it wrong.
GODAMNIT I HATE IT WHEN I"M RIGHT BUT PEOPLE WON"T LISTEN TO REASON!!!

I got the right ****ing answer! every other class in this damned school showing work only gives you partal credit if you get the wrong answer. but errr
