Question

Please also include the correct error and how you calculated it.

A binary mixture with a total mass of 0.810 ± 0.001 g of BaCl2•2H2O (molar mass = 244.27 g/mol) and Na2SO4 (molar mass = 142.04 g/mol) produces 0.466 ± 0.001 g of BaSO4 (molar mass = 233.39 g/mol); BaCl2•2H2O is the limiting reactant. The total mass of 0.810 ± 0.001 g contains both of the salts in unknown amounts. Calculate the percent of BaCl2•2H2O and the percent of Na2SO4 in the original mixture with error.

Answer 1

Answer

Reaction between BaCl₂.2H₂O and Na₂SO₄ taking place as follows-

BaCl₂.2H₂O + Na₂SO4 = BaSO₄ + 2NaCl + 2H₂O

Weight of BaSO₄ produced = 0.466 $\pm$ 0.001 g

Moles of BaSO₄ produced = 0.466/142.04 = 0.00199 moles.

So, in the above reaction, 0.00199 moles of BaCl₂.2H₂O react with 0.00199 moles of Na₂SO₄ to produce 0.00199 moles of BaSO₄.

So, weight of BaCl₂.2H₂O in the mixture = 0.00199 x 244.27 = 0.4860973 g

weight of Na₂SO₄ in the mixture = 0.00199 x 142.04 = 0.2826596 g

So, the percent of BaCl₂.2H₂O in the mixture = (0.4860973 x 100)/0.810 = 60.01 $\pm$ 0.001 %

the percent of Na₂SO₄ in the mixture = (0.2826596 x 100)/0.810 = 34.89 $\pm$ 0.001 %