Amount needed chemistry..

14.4 mg/L of HPO4

in the reaction: FeCl3 + HPO4 ---> FePO4 +3Cl + H

How much FeCl3 is needed ??

Can you help me get this started???

Thank you so much.

Several things. First, phosphoric acid is H3PO4, so the equation will be:

FeCl3 + H3PO4 ---> FePO4 +3 HCl

Now, you give a concentration of phosphoric acid (14.4 mg/L).

My next question is, did you really mean 14.4 mg/L? That is a very dilute solution of phosphoric acid. Might it actually have been 14.4 moles/L?

Again, though, somehow, you need to get to moles of H3PO4. Normally, the problem would include a volume. If that is the correct concentration (14.4 mg/L) you would multiply that concentration by the volume (in L) to get some mass of H3PO4. You would then divide that mass by the molar mass of H3PO4 to get to moles.

If the concentration was 14.4 mol/L, you still need to know a volume (in L) to get to moles of H3PO4.

Once you have gotten to moles H3PO4, you see from the equation that H3PO4 and FeCl3 react in a 1:1 ratio. So, the moles FeCl3 needed will be equal to the moles of H3PO4.

Finally, to get to a mass of FeCl3, you will multiply the moles FeCl3 needed by its molar mass to get to grams FeCl3.