What is the mass of 2.25 moles of manganese(IV) sulfide (MnS2)?The Periodic TableA.2.25 gB.1.35 × 1024 gC.3.63 × 1026 gD.268 g

To find the mass of 2.25 moles of manganese(IV) sulfide (MnS2), you need to know the molar mass of MnS2.

Step 1: Find the molar mass of MnS2. The molar mass of an element is its atomic weight in grams per mole (g/mol). According to the periodic table, the atomic weight of manganese (Mn) is about 54.94 g/mol and the atomic weight of sulfur (S) is about 32.07 g/mol.

Since MnS2 has one atom of Mn and two atoms of S, its molar mass is: 54.94 g/mol (for Mn) + 2 * 32.07 g/mol (for S) = 119.08 g/mol

Step 2: Multiply the number of moles by the molar mass to get the mass. 2.25 moles * 119.08 g/mol = 267.93 g

So, the mass of 2.25 moles of MnS2 is approximately 268 g. Therefore, the answer is D. 268 g.