Complete the following chemical formula and determine the final molarity of NaCl: __ BaCl2(aq) + __ Na2SO4(aq) --> +

You react 45g BaCl2 with an excess of Na2SO4 in 250mL of water. Find the molarity of NaCl produced.
A. 0.25
B. 0.58
C. 1.72
D. 3.82

[NaCl] = 1.72 M

Explanation:

First of all, we balance the equation:

BaCl₂ + Na₂SO₄ →  2NaCl + BaSO₄

We convert mass of barium chloride to moles

45 g . 1mol / 208.23g = 0.216 moles

Ratio is 1:2. This means, that our moles of reactant may produce the double of moles, of product.

0.216 . 2 = 0.432 moles of NaCl are been produced.

Molarity is mol/L. We convert volume of water from mL to L

250 mL . 1L /1000 mL = 0.250 L

[NaCl] = 0.432 mol/0.250L = 1.72 M