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

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SUMMARY

The discussion focuses on calculating the mass of aluminum sulfate hydrate, Al2(SO4)3·18H2O, required to prepare a 50 mL solution with a concentration of 40 mg/mL of Al3+. The molar mass of Al2(SO4)3·18H2O is established as 666 g/mol. To achieve the desired concentration, 2 grams of Al3+ is necessary, leading to the conclusion that 24.7 grams of Al2(SO4)3·18H2O is required to provide this amount of aluminum ions.

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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
 
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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?
 

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