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?

 

[Blueshift5]

of Al2(SO4)3.18H2O. To determine this, we first need to calculate the molar mass of Al2(SO4)3.18H2O by adding the molar masses of each element present:

- Aluminum (Al): 2 x 26.98 g/mol = 53.96 g/mol
- Sulfur (S): 3 x 32.06 g/mol = 96.18 g/mol
- Oxygen (O): 4 x 16.00 g/mol = 64.00 g/mol
- Hydrogen (H): 18 x 1.01 g/mol = 18.18 g/mol

Total molar mass of Al2(SO4)3.18H2O = 232.32 g/mol

Then, we can use the molar ratio between Al2(SO4)3.18H2O and Al3+ to determine the mass of Al2(SO4)3.18H2O needed to produce 2 g of Al3+:

2 g Al3+ x (1 mol Al2(SO4)3.18H2O / 3 mol Al3+) x (232.32 g/mol Al2(SO4)3.18H2O) = 24.7 g of Al2(SO4)3.18H2O

Therefore, to prepare 50 mL of a 40 mg/mL solution of Al3+, you will need 24.7 g of Al2(SO4)3.18H2O.