Question

1) Write the balanced molecular chemical equation for the reaction of 0.15 M barium (II) chloride, with 0.24 M potassium sulfate. You may need to consult Appendix E to determine the states of each reactant and product. Assume any insoluble products are completely insoluble.

2) write the balanced net ionic equation for the reaction above.

3) Determine the percent yield if 250.0 mL of each reactant were allowed to react, and a mass of 12.4567 g of solid were obtained. Be sure to also clearly identify the solid based on the reaction scheme above.

Answer 1

Answer -

Given,

0.15 M barium (II) chloride

0.24 M potassium sulfate

Volume of each reactant = 250.0 ml or 0.25 L

Mass of Solid obtained ()= 12.4567 g

1) Molecular Equation = ?

We know that, when barium (II) chloride react with potassium sulfate Barium Sulphate and Potassium Chloride is formed.So,

BaCl₂ (aq) + K₂SO₄ (aq) → BaSO₄ (s) + 2KCl (aq) [Balanced] [Answer]

2) Net ionic Equation = ?

Steps for writing Net ionic Equation are-

a. Write molecular equation.

b. write total ionic equation by spliting into ions.

solids can not be Split into ions.

c. Cancel the same ions on both side. And write Net Ionic Equation.

molecular equation - BaCl₂ (aq) + K₂SO₄ (aq) → BaSO₄ (s) + 2KCl (aq)

Total Ionic Equation - Ba²⁺(aq) + 2Cl^- (aq) + 2K⁺(aq)+ SO₄^2- (aq) → BaSO₄ (s) + 2K⁺(aq) + 2 Cl^- (aq)

Net ionic equation - Ba²⁺(aq) + SO₄^2- (aq) → BaSO₄ (s) [Answer]

3). percentage Yield = ?

We know that,

Molarity = Moles/ Volume in L

So, Moles = Molarity * Volume in L

Moles of BaCl₂ = 0.15 M * 0.25 L

Moles of BaCl₂ = 0.0375 mol

Moles of K₂SO₄ = 0.24 M * 0.25 L

Moles of K₂SO₄ = 0.06 mol

Moles of BaSO₄ = Mass/ Molar Mass = 12.4567 g/233.38 g/mol = 0.0534 mol

BaCl₂ (aq) + K₂SO₄ (aq) → BaSO₄ (s) + 2KCl (aq) [Balanced]

Using Stiochiometry,

It can be analysed that, for 1 mole of BaCl₂ , 1 mole K₂SO₄ is required. Because molar ratio in 1:1.

Moles of BaCl₂ is available in less amount so it is the limiting reagent

Also, For 1 mole of BaCl₂ , 1 mole of K₂SO₄

So, moles of K₂SO₄ produced theoretically is 0.0375 mol

Now,

Percentage Yield = (Actual Yield/Theoretical Yield) *100

Percentage Yield = (0.0534 mol/0.0375 mol) *100

Percentage Yield = 142.4% [Answer]