posted by Natasha .
500ml of 0.250M Na2SO4 solution is added to an aqueous solution of 15.00 g of BaCl2 resulting in the formation of white precipitate of BaSO4. How many moles and how many grams of BaSO4 are formed?
This is a limiting reagent problem.
Na2SO4 + BaCl2 ==> BaSO4 + 2NaCl
mols Na2SO4 = M x L = ?
mols BaCl2 = grams/molar mass = ?
Convert mols Na2SO4 to mols BaSO4 using the coefficients in the balanced equation.
Convert mols BaCl2 to mols BaSO4--same process.
It is likely you will get two different values for mols BaSO4. The correct value in limiting reagent problems is ALWAYS the smaller one and the reagent providing that value is the limiting reagent.
Then grams BaSO4 = mols BaSO4 (the smaller value) x molar mass BaSO4.