How many grams of manganese (IV) oxide are needed to make 5.6 Liters of a 2.1 solution? (solve for moles then convert to grams)

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

no

explanation:

go away, people want answers, not to hook up

the answer is d.) potassium