Question

how many ml of 0.218 m sodium sulfate react with exactly 25.34 ml of 0.113 m bacl₂ given the reaction: bacl₂(aq) + na₂so₄(aq) → baso₄(s) + 2nacl(aq)
options: 2.86, 13.1, 5.52, 24.6, none of the above

Explanation:

Step1: Calculate moles of BaCl₂

Moles = Molarity × Volume (L) = $0.113\ \text{M} × 0.02534\ \text{L} = 0.00286342\ \text{mol}$

Step2: Find moles of Na₂SO₄

Mole ratio BaCl₂:Na₂SO₄ = 1:1, so moles Na₂SO₄ = $0.00286342\ \text{mol}$

Step3: Calculate volume of Na₂SO₄

Volume (L) = Moles / Molarity = $\frac{0.00286342\ \text{mol}}{0.218\ \text{M}} ≈ 0.013135\ \text{L}$
Convert to mL: $0.013135\ \text{L} × 1000 = 13.135\ \text{mL} ≈ 13.1\ \text{mL}$

Answer:

B. 13.1