Calculate the mass of `BaCO_(3)` produced when excess `CO_(2)` is bubbled through a solution containing `0.205` moles of `Ba(OH)_(2)`.

Options

A. 40.5g
B. 48.5g
C. 4.5g
D. 60.5g

Answer

To calculate the mass of BaCO₃ produced when excess CO₂ is bubbled through a solution containing 0.205 moles of Ba(OH)₂, we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between barium hydroxide (Ba(OH)₂) and carbon dioxide (CO₂) can be represented as follows:

\[ \text{Ba(OH)}_2 + \text{CO}_2 \rightarrow \text{BaCO}_3 + \text{H}_2\text{O} \]

### Step 2: Determine the mole ratio
From the balanced equation, we can see that:
- 1 mole of Ba(OH)₂ reacts with 1 mole of CO₂ to produce 1 mole of BaCO₃ and 1 mole of H₂O.

### Step 3: Calculate the moles of BaCO₃ produced
Since we have 0.205 moles of Ba(OH)₂ and the reaction is a 1:1 ratio, it means that 0.205 moles of Ba(OH)₂ will produce 0.205 moles of BaCO₃.

### Step 4: Calculate the molar mass of BaCO₃
The molar mass of BaCO₃ can be calculated using the atomic masses:
- Barium (Ba) = 137.3 g/mol
- Carbon (C) = 12.0 g/mol
- Oxygen (O) = 16.0 g/mol (and there are 3 oxygen atoms)

Thus, the molar mass of BaCO₃ is:
\[ \text{Molar mass of BaCO}_3 = 137.3 + 12.0 + (3 \times 16.0) \]
\[ = 137.3 + 12.0 + 48.0 \]
\[ = 197.3 \text{ g/mol} \]

### Step 5: Calculate the mass of BaCO₃ produced
To find the mass of BaCO₃ produced, we multiply the moles of BaCO₃ by its molar mass:
\[ \text{Mass of BaCO}_3 = \text{moles of BaCO}_3 \times \text{molar mass of BaCO}_3 \]
\[ = 0.205 \text{ moles} \times 197.3 \text{ g/mol} \]
\[ = 40.5 \text{ g} \]

### Final Answer
The mass of BaCO₃ produced is **40.5 grams**.

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