A student mixes 50.0 mL of 1.00 M Ba(OH)2 with 83.5 mL of 0.475 M H2SO4. Calculate

9.26g of BaSO₄ are formed

Explanation:

The reaction of Ba(OH)₂ with H₂SO₄ is:

Ba(OH)₂ + H₂SO₄ → BaSO₄ + 2H₂O.

To solve this question we need to determine the moles of each reactant in order to find the limitng reactant. With the moles of limiting reactant we can find the theoretical moles of BaSO₄ produced and its mass, as follows:

Moles Ba(OH)₂:

50.0mL = 0.050L * (1.00mol / L) = 0.0500moles

Moles H₂SO₄:

83.5mL = 0.0835L * (0.475mol / L) = 0.0397moles

That means limiting reactant is H₂SO₄.

Based on the reaction, 1 mole of H₂SO₄ produce 1 mole of BaSO₄. The theoretical moles of BaSO₄ produced are 0.0397 moles.

The mass is (Molar mass BaSO₄: 233.38g/mol):

0.0397 moles BaSO₄ * (233.38g / mol) =