To solve the problem of how many grams of BaSO₄ are formed when 30 mL of 0.2 N BaCl₂ is mixed with 40 mL of 0.3 N Al₂(SO₄)₃, we will follow these steps:
### Step 1: Write the balanced chemical equation
The reaction between barium chloride and aluminum sulfate can be represented as:
\[ 3 \text{BaCl}_2 + \text{Al}_2(\text{SO}_4)_3 \rightarrow 3 \text{BaSO}_4 + 2 \text{AlCl}_3 \]
### Step 2: Calculate the milliequivalents of BaCl₂ and Al₂(SO₄)₃
Milliequivalents (mEq) can be calculated using the formula:
\[ \text{mEq} = \text{Normality} \times \text{Volume (mL)} \]
For BaCl₂:
\[ \text{mEq of BaCl}_2 = 0.2 \, \text{N} \times 30 \, \text{mL} = 6 \, \text{mEq} \]
For Al₂(SO₄)₃:
\[ \text{mEq of Al}_2(\text{SO}_4)_3 = 0.3 \, \text{N} \times 40 \, \text{mL} = 12 \, \text{mEq} \]
### Step 3: Determine the limiting reagent
According to the balanced equation, 3 mEq of BaCl₂ reacts with 1 mEq of Al₂(SO₄)₃. Therefore, we can find how many mEq of Al₂(SO₄)₃ will react with the available BaCl₂:
- From 6 mEq of BaCl₂, the amount of Al₂(SO₄)₃ required:
\[ \text{Required Al}_2(\text{SO}_4)_3 = \frac{6}{3} = 2 \, \text{mEq} \]
Since we have 12 mEq of Al₂(SO₄)₃ available, BaCl₂ is the limiting reagent.
### Step 4: Calculate the mEq of BaSO₄ formed
From the balanced equation, 3 mEq of BaCl₂ produces 3 mEq of BaSO₄. Therefore, 6 mEq of BaCl₂ will produce:
\[ \text{mEq of BaSO}_4 = 6 \, \text{mEq} \]
### Step 5: Calculate the equivalent weight of BaSO₄
The equivalent weight of BaSO₄ can be calculated as:
\[ \text{Equivalent weight} = \frac{\text{Molar mass}}{\text{n factor}} \]
The molar mass of BaSO₄ is approximately 233 g/mol, and since it dissociates to give 2 ions (Ba²⁺ and SO₄²⁻), the n factor is 2:
\[ \text{Equivalent weight of BaSO}_4 = \frac{233 \, \text{g/mol}}{2} = 116.5 \, \text{g/equiv} \]
### Step 6: Calculate the mass of BaSO₄ formed
Using the formula:
\[ \text{Weight} = \text{mEq} \times \text{Equivalent weight} / 1000 \]
Substituting the values:
\[ \text{Weight of BaSO}_4 = 6 \, \text{mEq} \times 116.5 \, \text{g/equiv} / 1000 = 0.699 \, \text{g} \]
### Final Answer
Thus, the mass of BaSO₄ formed is approximately **0.70 g**.
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