Step 1: Write the balanced chemical equation.
Na₂SO₄ + Ba(NO₃)₂ → 2NaNO₃ + BaSO₄
Step 2: Calculate the mass of BaSO₄ formed.
Based on the balanced chemical equation, 1 mole of Ba(NO₃)₂ is stoichiometrically equivalent to 1 mole of BaSO₄.
One molar (M) is equal to 1 mole per liter (mol/L).
25 mL = 0.025 L
The molar mass of BaSO₄ is 233.39 g/mol.
[tex]\begin{aligned} \text{mass of} \: \text{BaSO}_4 & = \text{0.025 L soln} \times \frac{\text{0.50 mol} \: \text{Ba}(\text{NO}_3)_2}{\text{1 L soln}} \times \frac{\text{1 mol} \: \text{BaSO}_4}{\text{1 mol} \: \text{Ba}(\text{NO}_3)_2} \times \frac{\text{233.39 g} \: \text{BaSO}_4}{\text{1 mol} \: \text{BaSO}_4} \\ & = \boxed{\text{2.9 g}} \end{aligned}[/tex]
Hence, the mass of solid BaSO₄ that will form when Na₂SO₄ reacts with 25 mL of 0.50-M Ba(NO₃)₂ is 2.9 g.
[tex]\\[/tex]
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