if 10.0 g Fe2(SO4)3 and 12.0g BaCl2 are reacted & 10.6 g of a solid is collected. calculate% yield.?

Question

the answer was given (79.1%) but i need helping getting to that point. thanks!!!!!!!!!!!

Answer 1

First, you have to identify what the solid is. Of the two possible products, the one that forms is barium sulfate, BaSO4 (FeCl3 is soluble). The net reaction is Ba(2+) + SO4(2-) --> BaSO4.

Next, you need to identify the limiting reactant, so you need to convert both masses you are given to moles. Ferric sulfate weighs 399.88 g/mole, while BaCl2 weighs 208.23 g/mole. So you have 0.0250 moles Fe2(SO4)3, which would produce 0.0750 moles of sulfate ion, and 0.0576 moles of barium chloride, which produces 0.0576 moles of barium ion. So barium is your limiting reactant, and the moles of barium sulfate you could form is 0.0576 moles, or 13.4 grams (MW BaSO4 is 233.39).

Since you formed 10.6, the percent yield is 10.6/13.4 X 100% or 79.1%.

Answer 2

yeild = 38.0g