Question

The amount of $BaSO _{4}$ formed upon mixing $100 \,mL$ of $20.8 \% BaCl _{2}$ solution with $50 \,mL$ of $9.8 \% H _{2} SO _{4}$ solution will be : $( Ba =137, Cl =35.5, S =32, H =1$ and $O =16)$

• A. 23.3 g
• B. 11.65 g
• C. 30.6 g
• D. 33.2 g

Answer 1

The Correct Option is B

To determine the amount of $BaSO_4$ formed from the reaction of $BaCl_2$ with $H_2SO_4$, we need to follow these steps:

1. Calculate the moles of $BaCl_2$:
   • The mass percentage of $BaCl_2$ solution is 20.8%.
   • This implies in 100 mL of the solution, 20.8 grams is $BaCl_2$.
   • Molar mass of $BaCl_2 = 137 + 2 \times 35.5 = 208 \, \text{g/mol}$.
   • Moles of $BaCl_2 = \frac{20.8}{208} = 0.1 \, \text{mol}$.
2. Calculate the moles of $H_2SO_4$:
   • The mass percentage of $H_2SO_4$ solution is 9.8%.
   • This implies in 50 mL of the solution, 9.8% of 50 grams is 4.9 \, \text{g}$ $H_2SO_4$ (assuming density near 1 g/mL).
   • Molar mass of $H_2SO_4 = 2 \times 1 + 32 + 4 \times 16 = 98 \, \text{g/mol}$.
   • Moles of $H_2SO_4 = \frac{4.9}{98} = 0.05 \, \text{mol}$.
3. Determine the limiting reagent:
   • The balanced chemical equation for the reaction is:
   $$\text{BaCl}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{BaSO}_4 + 2\text{HCl}$$
   • From the stoichiometry, 1 mole of $BaCl_2$ reacts with 1 mole of $H_2SO_4$ to produce 1 mole of $BaSO_4$.
   • Here, $H_2SO_4$ is the limiting reagent since we have only 0.05 mol available.
4. Calculate the mass of $BaSO_4$ formed:
   • Moles of $BaSO_4$ formed will be 0.05 mol (as $H_2SO_4$ is the limiting reagent).
   • Molar mass of $BaSO_4 = 137 + 32 + 4 \times 16 = 233 \, \text{g/mol}$.
   • Mass of $BaSO_4 = 0.05 \times 233 = 11.65 \, \text{g}$.

Therefore, the amount of $BaSO_4$ formed is 11.65 g.