The maximum amount of baso4 ppt on mixing equal volume of bacl2 [.5M] with h2so4 [1M] will correspond to

To determine the maximum amount of barium sulfate (BaSO4) precipitate formed when mixing equal volumes of barium chloride (BaCl2) and sulfuric acid (H2SO4), we need to analyze the stoichiometry of the reaction involved. The balanced chemical equation for the reaction is:

Understanding the Reaction

Barium chloride reacts with sulfuric acid to produce barium sulfate and hydrochloric acid:

BaCl2 + H2SO4 → BaSO4 (s) + 2 HCl

Identifying Reactants and Concentrations

In this scenario, you have:

• Barium chloride (BaCl2) at a concentration of 0.5 M
• Sulfuric acid (H2SO4) at a concentration of 1 M

Since we are mixing equal volumes of these solutions, let’s assume we are mixing 1 liter of each solution for simplicity. This means:

• Volume of BaCl2 = 1 L
• Volume of H2SO4 = 1 L

Calculating Moles of Each Reactant

Next, we calculate the number of moles of each reactant:

• Moles of BaCl2 = Concentration × Volume = 0.5 mol/L × 1 L = 0.5 moles
• Moles of H2SO4 = Concentration × Volume = 1 mol/L × 1 L = 1 mole

Determining the Limiting Reactant

From the balanced equation, we see that 1 mole of BaCl2 reacts with 1 mole of H2SO4 to produce 1 mole of BaSO4. Since we have 0.5 moles of BaCl2 and 1 mole of H2SO4, barium chloride is the limiting reactant. This means that the amount of BaSO4 formed will be determined by the amount of BaCl2 available.

Calculating the Amount of BaSO4 Precipitate

Since 1 mole of BaCl2 produces 1 mole of BaSO4, 0.5 moles of BaCl2 will produce 0.5 moles of BaSO4. To find the mass of BaSO4 precipitate, we need to know its molar mass:

• Molar mass of BaSO4 = 137.33 (Ba) + 32.07 (S) + 4 × 16.00 (O) = 233.39 g/mol

Now, we can calculate the mass of BaSO4:

• Mass of BaSO4 = Moles × Molar Mass = 0.5 moles × 233.39 g/mol = 116.695 g

Final Result

The maximum amount of barium sulfate precipitate that can be formed when mixing equal volumes of 0.5 M BaCl2 and 1 M H2SO4 is approximately 116.7 grams.