Question

If 5.5 moles of potassium permanganate (KMnO4) react in excess HCl, how much water, in grams, does the reaction produce? (Remember to balance the equation first.)

Answer 1

396 g H2O or ≈400 g H2O

Step-by-step explanation:

2KMnO4 + 16HCl ---> 2KCl +2 MnCl2 + 8H2O + 5 Cl2

2 mol 8 mol

5.5 mol x mol

x = 5.5*8/2 = 22 mol H2O

M(H2O) = 2*1.0 + 16.0 = 18.0 g/mol

22 mol H2O * (18.0 g H2O/1 mol H2O) = 396 g H2O or ≈400 g H2O