A 0.670 g sample of impure Al2(SO4)3 reacts with excess BaCl2. If the sample produces 0.760 g of BaSO4, what is the mass percent of Al2(SO4)3 in the sample?
Al2(SO4)3 + 3BaCl2 ==> 3BaSO4 + 2AlCl3
How many mols BaSO4 were collected? That's 0.760/molar mass BaSO4 = approx 0.0033 but you need to do it more accurately.
Using the coefficients in the balanced equation convert mols BaSO4 to mols Al2(SO4)3.
Convert mols Al2(SO4)3 to grams. g \ mols x molar mass
Then % Al2(SO4)3 = [g Al2(SO4)3/mass sample)]8100 = ?
To find the mass percent of Al2(SO4)3 in the sample, we need to determine the mass of Al2(SO4)3 in the sample and divide it by the total mass of the sample, then multiply by 100.
First, let's calculate the moles of BaSO4 that are produced. We can do this by dividing the mass of BaSO4 by its molar mass.
Molar mass of BaSO4 = (1 * atomic mass of Ba) + (1 * atomic mass of S) + (4 * atomic mass of O)
= (1 * 137.33 g/mol) + (1 * 32.07 g/mol) + (4 * 16.00 g/mol)
= 137.33 g/mol + 32.07 g/mol + 64.00 g/mol
= 233.40 g/mol
Moles of BaSO4 = mass of BaSO4 / molar mass of BaSO4
= 0.760 g / 233.40 g/mol
= 0.003255 mol
Since Al2(SO4)3 and BaSO4 have a 1:3 molar ratio according to the balanced equation, the moles of Al2(SO4)3 can be calculated as follows:
Moles of Al2(SO4)3 = 3 * moles of BaSO4
= 3 * 0.003255 mol
= 0.009765 mol
Now, let's calculate the mass of Al2(SO4)3 using its molar mass:
Molar mass of Al2(SO4)3 = (2 * atomic mass of Al) + (3 * atomic mass of S) + (12 * atomic mass of O)
= (2 * 26.98 g/mol) + (3 * 32.07 g/mol) + (12 * 16.00 g/mol)
= 53.96 g/mol + 96.21 g/mol + 192.00 g/mol
= 342.17 g/mol
Mass of Al2(SO4)3 = moles of Al2(SO4)3 * molar mass of Al2(SO4)3
= 0.009765 mol * 342.17 g/mol
= 3.343 g
Finally, we can calculate the mass percent of Al2(SO4)3 in the sample:
Mass percent of Al2(SO4)3 = (mass of Al2(SO4)3 / mass of sample) * 100
= (3.343 g / 0.670 g) * 100
= 498.81%
Therefore, the mass percent of Al2(SO4)3 in the sample is approximately 498.81%.