Question

A mixture contains 0.250 Mol of Mn2O3 and 20.0 g of MnO2.

1) How many atoms of oxygen are present in the mixture?

2) How many grams of manganese are in the mixture?

Answer 1

Mn2O3 moles = 0.250

MnO2 moles = mass /molar mass

                     = 20.0 / 87

                     = 0.230

total moles of oxygen in the mixture = 3 x 0.250 + 2 x 0.230

                                                           = 1.21

1mol contain -----------------------> 6.023 x 10^23 atoms

1.21 moles -------------------------> 1.21 x 6.023 x 10^23 atoms = 7.29 x 10^23 atoms

oxygen atoms = 7.29 x 10^23 atoms

2)

manganese moles = 2 x 0.250 + 0.230 = 0.73

molar mass of Mn = 54.93 g/mol

mass of manganese = 0.73 x 54.93 = 40.1 g

mass of manganese = 40.1 g