We are Given:
Final volume of the solution = 5.6 L
Concentration of the solution = 2.1 M
Moles of Manganese (IV) Oxide required:
Molar mass of Manganese (IV) Oxide = 87 grams/mol
since the concentration is 2.1 M, we will need 2.1 moles of Manganese Oxide for every 1L of the solution
Number of moles required = Molarity * Volume required
Number of moles required = 2.1 * 5.6 = 11.76 moles
Mass of Manganese (IV) Oxide required:
mass = molar mass * number of moles
Mass = 11.76 * 87
Mass = 1023.12 grams of Manganese (IV) Oxide
