Asked by lizzie

How many moles of O are present in 14.8 moles of Fe(NO3)3? Can you show work so I can understaned how to do.
Answered by bobpursley
If you have 14.8 moles of IronIII nitrate, you have 9*14.8 moles of O (count nine O in the formula). You also have three moles of N, and 14.8 moles of Fe.
Answered by Anonymous
im sorry i still do not understand
Answered by DrBob222
Look at the formula. If you have 1 mol iron(III)nitrate, [Fe(NO3)3], you have 1 mol Fe, 3 mols N, and 9 mols O. If you have 2 mols Fe(NO3)3, then you have 2*1 mol Fe, 2*3 mols N, and 2*9 mols O. So 14.8 mols Fe(NO3)3 will give you 14.8*1 mol Fe, 14.8*3 mols N, and 14.8*9 mols O.
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