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.
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.
im sorry i still do not understand
Well, let's break it down, or should I say, "mole it down"? Fe(NO3)3 means there are 3 nitrate ions per formula unit. Each nitrate ion contains 1 nitrogen atom (N) and 3 oxygen atoms (O). So, for every formula unit of Fe(NO3)3, we have 3 moles of O.
Therefore, for 14.8 moles of Fe(NO3)3, we would have 14.8 moles times 3 moles of O per 1 mole of Fe(NO3)3, giving us a grand total of... *drumroll*... 44.4 moles of O!
It's like a mole party in there, oxygen!
To determine the number of moles of oxygen (O) present in 14.8 moles of Fe(NO3)3, we need to analyze the chemical formula of Fe(NO3)3 to find the number of moles of oxygen atoms.
The chemical formula Fe(NO3)3 suggests that for each formula unit of Fe(NO3)3, there are three nitrate ions (NO3-) and each nitrate ion contains one nitrogen atom and three oxygen atoms.
First, let's calculate the moles of nitrate ions (NO3-) in 14.8 moles of Fe(NO3)3. Since one formula unit of Fe(NO3)3 contains three nitrate ions, we can use the mole-to-mole ratio:
Moles of NO3- = 14.8 moles Fe(NO3)3 × (3 moles NO3-/1 mole Fe(NO3)3)
= 14.8 × 3
= 44.4 moles NO3-
Now that we know the moles of nitrate ions, we can determine the moles of oxygen atoms. Since each nitrate ion contains three oxygen atoms, we again use the mole-to-mole ratio:
Moles of O = 44.4 moles NO3- × (3 moles O/1 mole NO3-)
= 44.4 × 3
= 133.2 moles O
Therefore, there are 133.2 moles of oxygen atoms present in 14.8 moles of Fe(NO3)3.