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
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