How do you do this?? Chemistry Help!!`?

Question

For this equation: Al2(SO4)3 + BaCl2 > BaSO4 + AlCl3

a .620g sample of impure Al2(SO4)3 reacts with excess BaCl2. If the sample produces .7g of BaSO4, what is the mass percent ofAl2(SO4)3 in the sample??

How do you do this? Can some one show me the steps?

Answer 1

Balance the eqn: 3 moles of BaSO4 per mole of Al2(SO4)3

grams of Al2(SO4)3 = 3*0.7/MW of BaSO4

% Al2(SO4)3 = 100 * g Al2(SO4)3/0.620

remember that there can only be 1 significant digit as the BaSO4 value was only 0.7!

Answer 2

First you have to balance the reaction:

Al2(SO4)3 + 3BaCl2 --> 3BaSO4 + 2AlCl3

So you know that 1 mole of Al2(SO4)3 is needed to obtain 3 moles of BaSO4. Now calculate how many moles are contained in 0,7 g of BaSO4 : number moles = mass / molar mass --> 0,7 / 233,4 = 0,0029991 mol.

Since you need 3 moles of BaSO4 to get 1 of Al2(SO4)3, you have to divide 0,0029991 per 3 to obtain 0,0009997 that is the number of moles of Al2(SO4)3 needed to produce 0,0029991 moles of BaSO4.

The simple proportion is 3 : 0,0029etc. = 1 : X --> X = 0,0009etc

Now you can reckon the mass of aluminium needed to carry on the reaction: molar mass x number moles = mass --> 342,17 x 0,0009997 = 0,342 g.

Just make the proportion 0,62 : 100 = 0,342 : X --> X = 55,16 %

I hope I've been clear...^^