5 moles of `Ba(OH)_2` are treated with excess of `CO_2` . How much `BaCO_2` will be formed?

Options

A. 39.4 g
B. 197 g
C. 591 g
D. 985 g

Explanation

To determine how much barium carbonate (BaCO₃) will be formed when 5 moles of barium hydroxide (Ba(OH)₂) are treated with excess carbon dioxide (CO₂), we can follow these steps:

### Step 1: Write the balanced chemical equation.
The reaction between barium hydroxide and carbon dioxide can be represented by the following balanced equation:

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

### Step 2: Identify the mole ratio.
From the balanced equation, we can see that 1 mole of barium hydroxide reacts with 1 mole of carbon dioxide to produce 1 mole of barium carbonate and 1 mole of water. Therefore, the mole ratio is:

- 1 mole of Ba(OH)₂ produces 1 mole of BaCO₃.

### Step 3: Calculate the amount of BaCO₃ produced.
Since we have 5 moles of Ba(OH)₂ and the reaction produces BaCO₃ in a 1:1 ratio, we can conclude that:

\[ \text{Moles of BaCO}_3 = \text{Moles of Ba(OH)}_2 = 5 \text{ moles} \]

### Conclusion:
Thus, when 5 moles of barium hydroxide are treated with excess carbon dioxide, 5 moles of barium carbonate will be formed.