Mass of precipitate formed when 3.75 L of 0.0570 M Ba(OH)2 is mixed with 3.25 L of 0.0782 M Na2SO4

Question

When 3.75 L of 0.0570 M Ba(OH)₂ is mixed with 3.25 L of 0.0782 M Na₂SO₄ a precipitate is formed. Write a balanced equation for this reaction (include the state of the reactants and products) and calculate the mass of the precipitate formed. Molar mass of BaSO₄ = 233.37 g/mol.

Answer 1

Write the correctly balanced equation:

Ba(OH)₂(aq) + Na₂SO₄(aq) ==> BaSO₄(s) + 2NaOH(aq) .. balanced equation

Given amounts of both reactants we much find which one is limiting. We can do this by dividing moles of each by the corresponding coefficient in the balanced equation.

For Ba(OH)₂: 3.75 L x 0.0570 mol / L = 0.21375 mols (÷1->0.21)

For Na₂SO₄: 3.25 L x 0.0782 mol / L = 0.2542 mols (÷1->0.25

Since 0.21 is less than 0.25, Ba(OH)₂ is the limiting reactant, and moles of Ba(OH)₂ will determine how much precipitate (BaSO₄) will form.

0.21375 mols Ba(OH)₂ x 1 mol BaSO₄ / mol Ba(OH)₂ x 233.37 g BaSO₄ / mol = 49.9 g BaSO₄ formed