The 500 cubic centimeter of 0.250 M Na2SO4 solution, added to an aqueous solution of 15.00 grams of barium
chloride, resulted in the formation of a white precipitate of
barium sulfate. How many moles and how many grams of
barium sulfate are formed, respectively?

0.072 moles, 16.776g.

Explanation:

The reaction of Na2SO4 and BaCl2 occurs as follows:

Na2SO4(aq) + BaCl2(aq) → BaSO4(s) + 2NaCl(aq)

To solve this question we must find the moles of each reactant. As the reaction is 1:1, the reactant with the low number of moles is limiting reactant. The moles of limiting reactant = Moles BaSO4. The mass can be obtained with the molar mass of BaSO4 -233.38g/mol-

Moles Na2SO4:

500cm³ = 0.500L * (0.250mol / L) = 0.125 moles

Moles BaCl2 -Molar mass: 208.23g/mol-

15.00g * (1mol / 208.23g) = 0.072 moles

The moles of BaSO4 are 0.072 moles and its mass is:

0.072 moles * (233.38g / mol) = 16.8g ≈ 16.776g

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