If 0.5 mol of `BaCl_(2)` is mixed with 0.2 mol of `Na_(3)PO_(4)`, the maximum number of moles of `Ba_(3)(PO_(4))_(2)` that can be formed is

Options

A. `0.7`
B. `0.5`
C. `0.1`
D. `0.2`

Explanation

To determine the maximum number of moles of \( \text{Ba}_3(\text{PO}_4)_2 \) that can be formed when mixing 0.5 moles of \( \text{BaCl}_2 \) with 0.2 moles of \( \text{Na}_3\text{PO}_4 \), we need to follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction is:
\[ 3 \text{BaCl}_2 + 2 \text{Na}_3\text{PO}_4 \rightarrow 1 \text{Ba}_3(\text{PO}_4)_2 + 6 \text{NaCl} \]

### Step 2: Identify the stoichiometric ratios
From the balanced equation, we can see that:
- 3 moles of \( \text{BaCl}_2 \) react with 2 moles of \( \text{Na}_3\text{PO}_4 \) to produce 1 mole of \( \text{Ba}_3(\text{PO}_4)_2 \).

### Step 3: Calculate the limiting reagent
Given:
- 0.5 moles of \( \text{BaCl}_2 \)
- 0.2 moles of \( \text{Na}_3\text{PO}_4 \)

Now, we need to determine how many moles of \( \text{BaCl}_2 \) are required to react with 0.2 moles of \( \text{Na}_3\text{PO}_4 \):
- From the stoichiometry, 2 moles of \( \text{Na}_3\text{PO}_4 \) require 3 moles of \( \text{BaCl}_2 \).
- Therefore, for 0.2 moles of \( \text{Na}_3\text{PO}_4 \):
\[ \text{BaCl}_2 \text{ required} = \frac{3}{2} \times 0.2 = 0.3 \text{ moles} \]

Since we have 0.5 moles of \( \text{BaCl}_2 \), which is more than the 0.3 moles required, \( \text{Na}_3\text{PO}_4 \) is the limiting reagent.

### Step 4: Calculate the maximum moles of \( \text{Ba}_3(\text{PO}_4)_2 \) produced
Using the stoichiometric ratios from the balanced equation:
- 2 moles of \( \text{Na}_3\text{PO}_4 \) produce 1 mole of \( \text{Ba}_3(\text{PO}_4)_2 \).
- Therefore, 0.2 moles of \( \text{Na}_3\text{PO}_4 \) will produce:
\[ \text{Ba}_3(\text{PO}_4)_2 \text{ produced} = \frac{1}{2} \times 0.2 = 0.1 \text{ moles} \]

### Conclusion
The maximum number of moles of \( \text{Ba}_3(\text{PO}_4)_2 \) that can be formed is **0.1 moles**.