To solve the problem of how many grams of BaSO₄ are formed when 20 mL of 0.1 N BaCl₂ is mixed with 30 mL of 0.2 N Al₂(SO₄)₃, we will follow these steps:
### Step 1: Calculate the milliequivalents of BaCl₂
To find the milliequivalents (mEq) of BaCl₂, we use the formula:
\[
\text{mEq} = \text{Normality} \times \text{Volume (L)}
\]
Given:
- Normality of BaCl₂ = 0.1 N
- Volume of BaCl₂ = 20 mL = 0.020 L
Calculating:
\[
\text{mEq of BaCl₂} = 0.1 \, \text{N} \times 0.020 \, \text{L} = 0.002 \, \text{Eq} = 2 \, \text{mEq}
\]
### Step 2: Calculate the milliequivalents of Al₂(SO₄)₃
Using the same formula:
\[
\text{mEq} = \text{Normality} \times \text{Volume (L)}
\]
Given:
- Normality of Al₂(SO₄)₃ = 0.2 N
- Volume of Al₂(SO₄)₃ = 30 mL = 0.030 L
Calculating:
\[
\text{mEq of Al₂(SO₄)₃} = 0.2 \, \text{N} \times 0.030 \, \text{L} = 0.006 \, \text{Eq} = 6 \, \text{mEq}
\]
### Step 3: Determine the limiting reagent
The balanced reaction for the formation of BaSO₄ is:
\[
\text{BaCl}_2 + \text{Al}_2(\text{SO}_4)_3 \rightarrow \text{BaSO}_4 + \text{AlCl}_3
\]
From the stoichiometry of the reaction:
- 1 mEq of BaCl₂ reacts with 1 mEq of Al₂(SO₄)₃.
We have:
- 2 mEq of BaCl₂
- 6 mEq of Al₂(SO₄)₃
Since we need 6 mEq of BaCl₂ to react with 6 mEq of Al₂(SO₄)₃, but we only have 2 mEq of BaCl₂, BaCl₂ is the limiting reagent.
### Step 4: Calculate the mEq of BaSO₄ formed
Since BaCl₂ is the limiting reagent, the mEq of BaSO₄ formed will be equal to the mEq of BaCl₂:
\[
\text{mEq of BaSO}_4 = 2 \, \text{mEq}
\]
### Step 5: Calculate the equivalent weight of BaSO₄
The molar mass of BaSO₄ is calculated as follows:
- Atomic weight of Ba = 137 g/mol
- Atomic weight of S = 32 g/mol
- Atomic weight of O = 16 g/mol (4 O atoms)
Calculating the molar mass:
\[
\text{Molar mass of BaSO}_4 = 137 + 32 + (4 \times 16) = 137 + 32 + 64 = 233 \, \text{g/mol}
\]
The equivalent weight of BaSO₄ is the molar mass divided by the number of equivalents (which is 1 for BaSO₄):
\[
\text{Equivalent weight of BaSO}_4 = \frac{233 \, \text{g/mol}}{1} = 233 \, \text{g}
\]
### Step 6: Calculate the weight of BaSO₄ formed
Using the formula:
\[
\text{Weight} = \text{mEq} \times \text{Equivalent weight}
\]
Substituting the values:
\[
\text{Weight of BaSO}_4 = 2 \, \text{mEq} \times \frac{233 \, \text{g}}{1000} = 0.466 \, \text{g}
\]
### Final Answer
The weight of BaSO₄ formed is 0.466 grams.
