What volume of .750M FeCl2 is needed to completely react with 1.00*10^2 mL of .250M KMnO4? PRETTY PLEASE HELP!

Given reaction:

5FeCl2(aq)+KMnO4(aq)+8HCl(aq)------>

5FeCl3(aq)+MnCl2(aq)+KCl(aq)+4H20(l)
(It was too long so I had to break up the equation.

Can you please explain how you got the answer step by step.

Thank you so much!!

Hurr. Here you're only really concerned with the first part of the equation where you have FeCl2 and KMnO4

This kind of questions means you need to know the amount that is reacting in moles.
From the equation you can see that 5 moles of FeCl2 completely reacts with 1 mole of KMnO4.

So we're using 100mL of .25M KMnO4. This can be easily converted to number of moles.

Remember M stands for molar which is moles/litre, so you have a .25M solution which is .25moles per litre of solution.

You have .1L of solution, therefore you have .1L * .25mol/L = .025moles of KMnO4.

Now, as I mentioned above we know that 5 moles of FeCl2 reacts with every mole of KMnO4, so we need 5x the number of moles of FeCl2, so 5 * .025 = .125moles of FeCl2

Now we know concentration of FeCl2, and the amount, we just need a volume.

Molar = mol/volume so volume = mol/molar

V(FeCl2) = .125mol/.75M = .166L or 1.66*10^2mL of FeCl2

Just curiously, what level of education are you at?