`30 mL` of `0.1MBaCl_(2)` is mixed with `40 mL` of `0.2 M Al_(2)(SO_(4))_(3)`. What is the weight of `BaSO_(4)` formed?
`BaCl_(2)+Al_(2)(SO_(4))_(3)rarrBaSO_(4)+AlCl_(3)`

Options

A. 0.999 g
B. 0.699 g
C. 0.799 g
D. 0.99 g

Answer

To solve the problem of determining the weight of BaSO₄ formed when 30 mL of 0.1 M BaCl₂ is mixed with 40 mL of 0.2 M Al₂(SO₄)₃, we can follow these steps:

### Step 1: Write the Balanced Chemical Equation
The balanced reaction for the formation of BaSO₄ is:
\[ \text{BaCl}_2 + \text{Al}_2(\text{SO}_4)_3 \rightarrow \text{BaSO}_4 + \text{AlCl}_3 \]

### Step 2: Calculate the Moles of BaCl₂
To find the moles of BaCl₂, we use the formula:
\[ \text{Moles} = \text{Volume (L)} \times \text{Molarity (M)} \]

Given:
- Volume of BaCl₂ = 30 mL = 0.030 L
- Molarity of BaCl₂ = 0.1 M

Calculating the moles:
\[ \text{Moles of BaCl}_2 = 0.030 \, \text{L} \times 0.1 \, \text{M} = 0.003 \, \text{moles} \, (3 \, \text{mmoles}) \]

### Step 3: Calculate the Moles of Al₂(SO₄)₃
Using the same formula for Al₂(SO₄)₃:
Given:
- Volume of Al₂(SO₄)₃ = 40 mL = 0.040 L
- Molarity of Al₂(SO₄)₃ = 0.2 M

Calculating the moles:
\[ \text{Moles of Al}_2(\text{SO}_4)_3 = 0.040 \, \text{L} \times 0.2 \, \text{M} = 0.008 \, \text{moles} \, (8 \, \text{mmoles}) \]

### Step 4: Determine the Limiting Reagent
From the balanced equation, we see that:
- 1 mole of Al₂(SO₄)₃ reacts with 3 moles of BaCl₂.

Thus, for 0.008 moles of Al₂(SO₄)₃, the required moles of BaCl₂ would be:
\[ 0.008 \, \text{moles Al}_2(\text{SO}_4)_3 \times 3 = 0.024 \, \text{moles BaCl}_2 \]

Since we only have 0.003 moles of BaCl₂ available, BaCl₂ is the limiting reagent.

### Step 5: Calculate the Moles of BaSO₄ Formed
From the balanced equation, 1 mole of BaCl₂ produces 1 mole of BaSO₄. Therefore, the moles of BaSO₄ formed will be equal to the moles of BaCl₂:
\[ \text{Moles of BaSO}_4 = 0.003 \, \text{moles} \]

### Step 6: Calculate the Weight of BaSO₄
To find the weight of BaSO₄, we use the formula:
\[ \text{Weight} = \text{Moles} \times \text{Molar Mass} \]

The molar mass of BaSO₄ (Barium Sulfate) is approximately:
- Ba = 137 g/mol
- S = 32 g/mol
- O₄ = 16 g/mol × 4 = 64 g/mol
- Total = 137 + 32 + 64 = 233 g/mol

Calculating the weight:
\[ \text{Weight of BaSO}_4 = 0.003 \, \text{moles} \times 233 \, \text{g/mol} = 0.699 \, \text{grams} \]

### Final Answer
The weight of BaSO₄ formed is **0.699 grams**.

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