Al2(SO4)3(aq) + BaCl2(aq) BaSO4(s) + AlCl3(aq)
Consider the unbalanced equation above. A 0.680 g sample of impure Al2(SO4)3 reacts with excess BaCl2. If the sample produces 0.530 g of BaSO4, what is the mass percent of Al2(SO4)3 in the sample?
To solve this problem, we need to determine the balanced equation for the reaction, and then use stoichiometry to calculate the moles and mass of Al2(SO4)3. Finally, we can calculate the mass percent.
Step 1: Write the balanced equation
Al2(SO4)3(aq) + 3BaCl2(aq) -> 3BaSO4(s) + 2AlCl3(aq)
Step 2: Calculate the moles of BaSO4
Using the given mass of BaSO4 (0.530 g) and its molar mass (233.39 g/mol):
moles of BaSO4 = mass / molar mass = 0.530 g / 233.39 g/mol
Step 3: Calculate the moles of Al2(SO4)3
From the balanced equation, we can see that the mole ratio between BaSO4 and Al2(SO4)3 is 3:1. Therefore:
moles of Al2(SO4)3 = moles of BaSO4 / 3
Step 4: Calculate the mass of Al2(SO4)3
Using the moles of Al2(SO4)3 and its molar mass (342.15 g/mol):
mass of Al2(SO4)3 = moles of Al2(SO4)3 * molar mass = moles of Al2(SO4)3 * 342.15 g/mol
Step 5: Calculate the mass percent of Al2(SO4)3
mass percent of Al2(SO4)3 = (mass of Al2(SO4)3 / mass of impure sample) * 100
Now let's substitute the given values and solve the equation:
Step 2: moles of BaSO4 = 0.530 g / 233.39 g/mol = 0.00227 mol
Step 3: moles of Al2(SO4)3 = 0.00227 mol / 3 = 0.00075 mol
Step 4: mass of Al2(SO4)3 = 0.00075 mol * 342.15 g/mol = 0.256 g
Step 5: mass percent of Al2(SO4)3 = (0.256 g / 0.680 g) * 100 = 37.6%
Therefore, the mass percent of Al2(SO4)3 in the sample is 37.6%.