Calculate Al3+ Mass: Al2(SO4)3.18H2O for 40mg/mL Solution in 50mL

[fishingspree2]

Homework Statement

Find the required mass of Al₂(SO₄)₃.18H₂O (molar mass = 666 g/mol) to prepare 50 mL of an aqueous solution, concentration = 40 mg of Al³⁺ per mL

The Attempt at a Solution

0.04 grams of Al³⁺/1mL = x g / 50 mL
x = 2g

I think this is the required mass of Al³⁺, now I don't know how to find how much Al₂(SO₄)₃.18H₂O do I need to get 2 g of Al³⁺

The answer is 24.7g

 

[Borek]

How many moles of Al do you need?

In how many moles of aluminum sulfate there will be required amount of Al?

What will be mass of these moles?