Answer to Question #92076 in Organic Chemistry for Ummulkhairi

Question

What is the experimental yield in grams when the % yield is 10.82% when 80g of BaCl react in solution with excess sodium sulphate?

Answer 1

BaCl2 + Na2SO4 -> BaSO4 + 2NaCl

n = m/M

n(BaCl2) = 80/208.23 = 0.384 mol

As mole ratio n(BaCl2)=n(BaSO4), then n(BaSO4) = 0.384 mol

m(theoretical)(BaSO4) = M*n = 233.39*0.384 = 89.6 g

m(actual)(BaSO4) = 89.6 *0.1082 = 9.70 g

Answer: 9.70 g