Question

question 12 (1 point) retake question if 50 ml of h₂so₄ yields 0.300 g of baso₄ when reacts with excess of bacl₂, what is the concentration of h₂so₄ in mol/l bacl₂(aq) + h₂so₄(aq) ---> baso₄(s) + 2 hcl(aq) blank mol/l

Explanation:

Step1: Calculate moles of BaSO₄

Molar mass of \( \text{BaSO}_4 \) is \( 137.33 + 32.07 + 4\times16.00 = 233.40 \, \text{g/mol} \).
Moles of \( \text{BaSO}_4 = \frac{0.300 \, \text{g}}{233.40 \, \text{g/mol}} \approx 0.001285 \, \text{mol} \).

Step2: Relate moles of H₂SO₄ and BaSO₄

From the reaction \( \text{BaCl}_2 + \text{H}_2\text{SO}_4
ightarrow \text{BaSO}_4 + 2\text{HCl} \), the mole ratio of \( \text{H}_2\text{SO}_4 \) to \( \text{BaSO}_4 \) is \( 1:1 \). So moles of \( \text{H}_2\text{SO}_4 = 0.001285 \, \text{mol} \).

Step3: Calculate concentration of H₂SO₄

Volume of \( \text{H}_2\text{SO}_4 = 50 \, \text{mL} = 0.050 \, \text{L} \).
Concentration \( = \frac{\text{moles}}{\text{volume}} = \frac{0.001285 \, \text{mol}}{0.050 \, \text{L}} \approx 0.0257 \, \text{mol/L} \) (or more precisely, using exact calculation: \( \frac{0.300}{233.40 \times 0.050} \approx 0.0257 \)).

Answer:

\( \approx 0.0257 \) (or more accurately, \( \frac{0.300}{233.40 \times 0.050} \approx 0.0257 \) mol/L)