Answer :
The maximum amount of barium sulfate that can be formed is approximately 89.4 grams, and the amount of excess reagent (barium chloride) remaining after the reaction is complete is approximately 30.34 grams.
To find the maximum amount of barium sulfate that can be formed, we need to determine the limiting reagent in the reaction. The limiting reagent is the reactant that is completely consumed and determines the amount of product formed.
First, let's calculate the number of moles for each reactant;
Molar mass of barium chloride (BaCl₂) = 137.33 g/mol
Molar mass of potassium sulfate (K₂SO₄) = 174.26 g/mol
Moles of barium chloride = mass / molar mass
= 83.0 g / 137.33 g/mol ≈ 0.604 mol
Moles of potassium sulfate = mass / molar mass
= 66.9 g / 174.26 g/mol ≈ 0.383 mol
Next, we need to determine the stoichiometric ratio between the reactants and the product. From the balanced equation;
1 mole of barium chloride reacts with 1 mole of potassium sulfate to produce 1 mole of barium sulfate.
Since the ratio is 1:1, the limiting reagent will be the one with fewer moles. In this case, potassium sulfate is the limiting reagent because it has fewer moles (0.383 mol) compared to barium chloride (0.604 mol).
Now, let's calculate the maximum amount of barium sulfate that can be formed using the limiting reagent;
Molar mass of barium sulfate (BaSO₄) = 233.39 g/mol
Moles of barium sulfate formed = moles of limiting reagent (potassium sulfate) = 0.383 mol
Mass of barium sulfate formed = moles × molar mass
= 0.383 mol × 233.39 g/mol ≈ 89.4 g
Therefore, the maximum amount of barium sulfate that can be formed is approximately 89.4 grams.
To determine the amount of excess reagent remaining after the reaction is complete, we need to compare the moles of the excess reagent (barium chloride) with the moles used in the reaction.
Moles of excess reagent = moles of excess reagent - moles used in the reaction
Moles of excess reagent = 0.604 mol - 0.383 mol
= 0.221 mol
Now, let's calculate the mass of the excess reagent remaining;
Mass of excess reagent remaining = moles of excess reagent × molar mass
Mass of excess reagent remaining = 0.221 mol × 137.33 g/mol
≈ 30.34 g
Therefore, the amount of excess reagent (barium chloride) remaining after the reaction is complete is approximately 30.34 grams.
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