Calculate the limiting reagent and the theoretical yield in grams of barium sulfate?

Question

3 BaCl2 + Al2(SO4)3 ==> 2AlCl3+3BaSO4(s)

mm= 208.2 342.0

25.0 grams of barium chloride is reacted with 20.3 grams of aluminum sulfate

Answer 1

Assuming that the reaction goes to completion.

Compute for the number of moles per reactant given the molecular weight of each compound.

MW of BaCl2 = 208.2 g/mol

MW of Al2(SO4)3 = 342.0 g/mol

Moles of Reactant:

mol of BaCl2 = 25.0 / 208.2 = 0.120 mol

mol of Al2(SO4)3 = 20.3 / 342.0 = 0.060 mol

Based on the reaction, 3 BaCl2 per 1 Al2(SO4)3. Therefore, 0.120 / 3 = 0.040 mol of Al2(SO4)3 will be needed leaving 0.020 mol unreacted.  The limiting reactant is BaCl2.

MW of BaSO4 = 233.40 g/mol

Since 0.120 mol BaCl2 will react, 0.120 mol BaSO4 will also be the product.

0.120 mol (233.40 g/mol) = 28.0 g BaSO4 yield

0.020 mol of Al2(SO4)3 unreacted = 0.020 mol (342 g/mol) = 6.84 g Al2(SO4)3 unreacted