1. Write the balanced chemical equation:

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

2. Calculate the moles of each reactant:

* Moles of BaCl₂:

(22.6 mL) * (1 L / 1000 mL) * (0.160 mol/L) = 0.00362 mol BaCl₂

* Moles of Na₂SO₄:

(54.6 mL) * (1 L / 1000 mL) * (0.055 mol/L) = 0.00300 mol Na₂SO₄

3. Determine the limiting reactant:

* The balanced equation shows a 1:1 mole ratio between BaCl₂ and Na₂SO₄.

* Since we have fewer moles of Na₂SO₄ (0.00300 mol) than BaCl₂ (0.00362 mol), Na₂SO₄ is the limiting reactant. This means it will be completely consumed, and the amount of BaSO₄ formed will be determined by the amount of Na₂SO₄.

4. Calculate the moles of BaSO₄ formed:

* The 1:1 mole ratio from the balanced equation tells us that 0.00300 mol of Na₂SO₄ will produce 0.00300 mol of BaSO₄.

5. Convert moles of BaSO₄ to grams:

* Molar mass of BaSO₄ = 137.33 g/mol + 32.06 g/mol + (4 * 16.00 g/mol) = 233.39 g/mol

* Mass of BaSO₄ = (0.00300 mol) * (233.39 g/mol) = 0.700 g

Therefore, 0.700 grams of solid barium sulfate will form.