The question isn't exactly clear. It SAYS 690 mol O2 reacted (meaning that you had 690 + 163 = 853 mols initially). If that is so, the amount of O2 left is not relevant. So you would have 690 x (2 mols Al2O3/3 mols O2) = 460 mols Al2O3 formed.
The other interpretation is that we started with 690 and ended up with 163 which means 527 mols O2 actually reacted. In that case, the mols Al2O3 formed is
mols O2 reacted ==> 690 - 163 = 527.
Convert mols O2 reacted to mols Al2O3 formed by
527 x (2 mols Al2O3/3 mols O2) = 351.
Do you have an answer manual or do you know the answer? Take the interpretation that fits the answer.
A sample of 690. moles of O2 reacted according to the balanced equation,
4Al+3O2=2Al2O3
after reaction, 163 moles of O2 remained. calculate the moles of Al2O3 formed.
Ive tried this problem many times and i keep getting it wrong. please help! thanks
2 answers
thanks sooo much. it was the second solution u gave. thanks again
1 answer