How many moles of oxygen atoms are in 10.8 mol of iron(III) nitrate?

Question

Cameron Andrews · Accepted Answer

There are #"97.2 mol of O"# in #"10.8 mol of Fe"("NO"_3)_3#.

There are 9 atoms of #"O"# in 1 formula unit of #"Fe"("NO"_3)_3#, so there are #"9 mol of O"# in #"1 mol of Fe"("NO"_3)_3#. Your conversion factor is either #"9 mol O"/("1 mol Fe"("NO"_3")_3)# or #("1 mol Fe"("NO"_3)_3)/"9 mol O"#. To get the desired units on the answer, you select the one that cancels units. #10.8 color(red)(cancel(color(black)("mol Fe"("NO"_3)_3))) × "9 mol O"/(1 color(red)(cancel(color(black)("mol Fe"("NO"_3)_3)))) = "97.2 mol O"#

Luna Chen · Answer

undefined To find the number of moles of oxygen atoms in 10.8 moles of iron(III) nitrate (Fe(NO3)3), you first need to determine the ratio of oxygen atoms to iron(III) nitrate molecules in the compound. In iron(III) nitrate, there are three nitrate ions (NO3-) per iron ion (Fe^3+), and each nitrate ion contains three oxygen atoms.

So, for each iron(III) nitrate molecule, there are $3 \times 3 = 9$ oxygen atoms.

Now, you can calculate the number of moles of oxygen atoms in 10.8 moles of iron(III) nitrate:

$$ \text{Number of moles of oxygen atoms} = 10.8 \text{ moles} \times 9 $$

$$ = 97.2 \text{ moles} $$

Therefore, there are 97.2 moles of oxygen atoms in 10.8 moles of iron(III) nitrate.