How many grams of iron (III) sulfate are in 225 mL of 0.500 M solution of iron(III)sulfate?

Answer:

45 g

Step-by-step explanation

Iron(III) sulfate - Fe₂(SO₄)₃

Moles of iron(III) sulfate =Molarity×Volume(L)

=0.500 M×0.225 L

=0.1125 mol

Molar mass of Fe₂(SO₄)₃ =2×56+3×96 =400 g/mol

Mass of Fe₂(SO₄)₃ =Moles×Molar mass

=0.1125 mol×400 g/mol

=45 g