How many moles of MnCl2 can be produced when 25 g KMnO4 are mixed with 85 g HCl

Answer 1

2KMnO₄ + 16HCl ==> 2MnCl₂ + 2KCl + 5Cl₂ + 8H₂O ... balanced equation

25 g KMnO4 x 1 mol/158 g x 2 mol MnCl2/2 mol KMnO4 x 126 g/mol MnCl2 = 19.9 g MnCl2

85 g HCl x 1 mol/36.5 g x 2 mol MnCl2/16 mol HCl x 126 g/mol MnCl2 = 36.7 g MnCl2

Since KMnO4 is limiting, 19.9 g MnCl2 would be produced, theoretically.