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

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.

thanks sooo much. it was the second solution u gave. thanks again

Sure, I'll be happy to guide you through the steps of solving this problem and help you find the correct answer.

First, let's analyze the given information:

- Initial moles of O2 = 690 moles
- Moles of O2 remaining after the reaction = 163 moles

To find the moles of Al2O3 formed, we need to determine the change in moles of O2. This can be done by finding the difference between the initial moles of O2 and the moles of O2 remaining.

Change in moles of O2 = Initial moles of O2 - Moles of O2 remaining
Change in moles of O2 = 690 moles - 163 moles
Change in moles of O2 = 527 moles

According to the balanced equation, the ratio of O2 to Al2O3 is 3:2. This means that for every 3 moles of O2 consumed, 2 moles of Al2O3 are formed.

To find the moles of Al2O3 formed, we can set up a proportion using the ratio from the balanced equation:

(Change in moles of O2) / (moles of O2 consumed per mole of Al2O3) = (moles of Al2O3 formed)

(527 moles O2) / (3 moles O2) = (moles of Al2O3 formed) / (2 moles Al2O3)

To solve for the moles of Al2O3 formed, we can cross-multiply and divide:

(527 moles O2 * 2 moles Al2O3) / (3 moles O2) = moles of Al2O3 formed
(1054 moles Al2O3) / (3 moles O2) = moles of Al2O3 formed
≈ 351.33 moles of Al2O3 formed (rounded to two decimal places)

Therefore, the moles of Al2O3 formed is approximately 351.33 moles.