chem help?

Question

3.45 grams of iron (ii) chloride was mixed with potassium permangante and hydrochloric acid to yield iron (iii) chloride, manganese (iv) chloride, potassium chloride, and water. how many moles of hydrochloric acid were produced and how many grams manganese (iv) chloride were made in the reaction.?


please help..

Answer 1

First u have to write the net ionic reaction, which will help u find the balanced equation. This will take too long to explain.

3FeCl2 + MnO4 + 8HCl--> 3FeCl3 + MnCl4 + 4H2O

3.45g FeCl2 x 1 MnCl4 x 195g
_________________________ = 1.78 g MnCl4
126g/mol.........3 FeCl2.....mol


(126 is the MM of FeCl2 and 195 is the MM of MnCl4)

for the other part H+ wasn't produced, water was. if they asked for water...

3.45g FeCl2 x 4 H2O
__________________=0.037mol H2O
126g/mol.........3 FeCl2

Answer 2

You have the balanced equation
3FeCl2 + KMnO4 + 8HCl ---> 3FeCl3 + MnCl4 + KCl + 4H2O

Convert 3.45 g iron(II) chloride into moles and you get 0.0272 mols.
Therefore, you need 0.0726 mols HCl for the reactions and 0.00907 mols MnCl4.
Convert MnCl4 into grams and you get 1.785 grams of Manganese chloride