Answer
3.265x10^22 atoms O
Work Step by Step
First we need to know how many moles of O are present in 1 mole of Al(NO3)3. Looking at the formula we see that there are 9: 3 in each NO3 and there are 3 NO3s so 3*3 = 9.
Then, 1 mole of Al(NO3)3 would have 6.022 x 10^23 atoms of Al(NO3)3 so then
6.025x10^-3 mol * (6.022 x 10^23 molecules Al(NO3)3 /1mole)*(9 atoms O/ 1 molecule Al(NO3)3) = 3.265x10^22 atoms O
