What is the mass in grams of 1.223 mol of iron (iii) sulfate?

The answer is

mass = 489g

The explanation:

1- iron (III) sulfate chemical formula is

Fe2(SO4)3

2- the molar mass of Fe2(SO4)3 is

= 55.8*2 + 32*3 + (16*4)*3 =399.9 g/mol

3- To get the mass we can use this formula:

Mass = moles * molar mass

= 1.223 mol * 399.9 g/mol

= 489 g

∴ the mass in grams of 1.223 mol iron(iii) sulfate = 489g