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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.)

If 55 moles of potassium permanganate KMnO4 react in excess HCl how much water in grams does the reaction produce Remember to balance the equation first class=

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SvetkaChem

Answer:

396 g H2O or  ≈400 g H2O

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