Correct each of the following formulas:

a. Barium oxide is$\mathit{B}\mathit{a}{\mathit{O}}_{\mathbf{2}}$

b. Iron(II) nitrate is$\mathit{F}\mathit{e}{\mathbf{\left(}\mathbf{N}{\mathbf{O}}_{\mathbf{3}}\mathbf{\right)}}_{\mathbf{3}}$

c. Magnesium sulfide is$\mathit{M}\mathit{n}\mathit{S}{\mathit{O}}_{\mathbf{3}}$

Barium has a charge of +2 and oxide has a charge of -2.

The charges of each ion will just interchange as subscripts of its molecular formula.

The subscript should also be reduced to the smallest whole numbers that retain the ratio of the atoms.

Barium oxide is$B{a}^{2+}+O^{2-}\to BaO$ .

Iron has a charge of +2 and nitrate has a charge of -1.

The charges of each ion will just interchange as subscripts of its molecular formula

Iron (II) nitrate is$F{e}^{2+}N{O}_3^{1-}\to Fe{\left(N{O}_3\right)}_2$ .

Magnesium has a charge of +2 and sulfide has a charge of - 2.

The charges of each ion will just interchange as subscripts of its molecular formula

The subscript should also be reduced to the smallest whole numbers that retain the ratio of the atoms.

Magnesium sulfide is $M{g}^{2+}+S^{2-}\to MgS$.