Question

I need a hand with this one:

Having this equation:
Na2SO4 +BaCl2 = BaSO4+NaCl
If 10.09 grams of Na2SO4 were added to a 50mL barium solution and reacted completely, what was the BaCl2 molarity of the solution?
Yes, I already know its a displacement reaction

Thx

Answer 1

[BaCl₂] = 1.42M

Step-by-step explanation:

Given: Na₂SO₄(aq) + BaCl₂(aq) => 2NaCl(aq) + BaSO₄(s)

The balanced equation shows a 1:1 reaction between Na₂SO₄(aq) & BaCl₂(aq) and means the moles of sodium sulfate used equals the moles of baium chloride used. The concentration is the theoretical mole value calculated divided by 0.050 Liters.

Na₂SO₄(aq) + BaCl₂(aq) => 2NaCl(aq) + BaSO₄(s)

Moles of Na₂SO₄(aq) = moles of BaCl₂(aq) = 10.09g/142.04g/mol = 0.071mole

∴ Molar Concentration of BaCl₂(aq) = moles BaCl₂/Volume of Soln in Liters

= 0.071mole/0.050Liters = 1.42M in BaCl₂.