An aqueous barium chloride solution was added to a 15.0 mL sample of a sodium sulfate solution according to the reaction below. If 1.83 grams of solid were recovered, what was the molarity of the original sodium sulfate solution?

Question

contestada

Respuesta :

nkirotedoreen46 nkirotedoreen46 · Answer 1 · 2019-10-26T02:34:09+02:00

0.523 M

The reaction between barium chloride and sodium sulfate is given by

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

We are given,

Volume of Na₂SO₄ = 15.0 mL

Mass of of solid BaSO₄ = 1.83 g

Required to determine the molarity of Na₂SO₄ solution

we will use the following steps

Mass of BaSO₄ = 1.83 g

To get the number of moles we divide mass by the molar mass

Molar mass of BaSO₄ = 233.38 g/mol

Number of moles = 1.83 g ÷ 233.38 g/mol

= 0.00784 moles

From the balanced equation for every 1 mole of sodium sulfate used 1 mole of BaSO₄ was produced.

Therefore, the mole ratio of Na₂SO₄ : BaSO₄ is 1 : 1

Hence, moles of Na₂SO₄ will also be 0.00784 moles

Volume of sodium sulfate = 15.0 mL or 0.015 L

Number of moles = 0.00784 moles

But, molarity = Moles ÷ Volume

= 0.00784 moles ÷ 0.015 L

= 0.5227 M

= 0.523 M

Thus, the molarity of original sodium sulfate is 0.523 M