Question

A student performed a precipitation reaction using 0.527 g of BaCl2 dissolved in water with an
excess of Na2SO4. If 0.551 g of BaSO4 is recovered, what was the percent yield of BaSO4?

Answer 1

mp (BaSO4) = 0,551 g
m (BaCl2) = 0,527 g
η(BaSO4) - ?
Na2SO4+ BaCl2 = 2NaCl + BaSO4↓
nt (BaSO4)= n (BaCl2) = m(BaCl2)/ M(BaCl2)= 0,527g/ 208g/mol = 0,003mol.
mt(BaSO4) = nt(BaSO4)*M (BaSO4)= 0,003 mol * 233 g/mol = 0,699g.
η(BaSO4)= mp(BaSO4) /mt(BaSO4) *100%=0,551g / 0,699g *100%= 78,82%