When amounts are given for BOTH reactants you know the question is a limiting reagent (LR) problem.
3BaCl2 + Fe2(SO4)3 ==> 3BaSO4 + 2FeCl3
millimols BaCl2 = mL x M = 100 x 0.1 = 10
mmols Fe3(SO4)3 = 100 x 0.1 = 10
How much BaSO4 can be formed from the BaCl2 IF we had all of the Fe3(SO4)3 we needed.
That's 10 mmols BaCl2 x (3 mol BaSO4/3 mols BaCl2) = 10 x 1/1 = 10 mmols BaSO4
How much BaSO4 can be formed from the Fe3(SO4)2 IF we had all of the BaSO4 needed.
That's 10 mmols Fe2(SO4)3 x (3 mols BaSO4/1 mol Fe2(SO4)3) = 10 x 3/1 = 30
You see the number for BaSO4 is different which means one of them is not right; the correct value in LR problems is ALWAYS the smaller value and the reagent producing that value is the LR. So we will produce 10 mmoles (0.01 mols) BaSO4 and use all of the BaCl2. Convert 0.01 mols BaSO4 to grams. g = mols x molar mass = ?
what mass of barium sulfate can be produced when 100 ml 0.100 M of solution of barium chloride is mixed with 100.0 ml of a 0.100 M solution of Iron (III) sulfate
1 answer
1 answer