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) a/ mol/l

Explanation:

Step1: Calculate moles of \( BaSO_4 \)

Molar mass of \( BaSO_4 \) is \( 137.33 + 32.07 + 4\times16.00 = 233.40 \, g/mol \).
Moles of \( BaSO_4 = \frac{mass}{molar\ mass} = \frac{0.300\ g}{233.40\ g/mol} \approx 0.001285\ mol \).

Step2: Relate moles of \( H_2SO_4 \) and \( BaSO_4 \)

From the reaction: \( 1\ mol\ H_2SO_4
ightarrow 1\ mol\ BaSO_4 \).
Thus, moles of \( H_2SO_4 = 0.001285\ mol \).

Step3: Calculate concentration of \( H_2SO_4 \)

Volume of \( H_2SO_4 = 50\ mL = 0.050\ L \).
Concentration \( (M) = \frac{moles}{volume} = \frac{0.001285\ mol}{0.050\ L} \approx 0.0257\ mol/L \).

Answer:

\( \approx 0.0257 \) (or more precisely, using exact calculation: \( \frac{0.300}{233.40\times0.050} \approx 0.0257 \))