Chem help plz!!!

Question

Iron (II) can be precipitated from neutral solutions by oxidizing it with MnO₄^–:

MnO−4​(aq)+3Fe(OH)+(aq)+2OH−(aq)+2H2​O(l)MnO2​(s)+3Fe(OH)3​(s)

 What volume of 0.825 M MnO₄^– is needed to completely react with 50.0 mL of 0.532 M Fe(II)?

Answer 1

Using the stoichiometry, the approach is to start with what you know (an amount of stuff) and solve for an amount:

50 ml Iron soln (1 liter/1000ml)(.532 moles Fe(II)/liter)(1 PM/3Fe(II))(1 liter/ .825 mole PM)(1000 ml/liter) = ans in ml of PM sol'n (PM stands for permanganate)

Note that you can ignore the ml-->l and back steps when you do calculation.