If 0.5 mole of `BaCl_(2)` is mixed with 0.20 mole of `Na_(3)PO_(4)`, the maximum number of `Ba_(3)(PO_(4))_(2)` that can be formed is :

Options

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

Explanation

To solve the problem, we need to determine how many moles of barium phosphate, \( \text{Ba}_3(\text{PO}_4)_2 \), can be formed when 0.5 moles of barium chloride, \( \text{BaCl}_2 \), are reacted with 0.2 moles of sodium phosphate, \( \text{Na}_3\text{PO}_4 \).

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equation:**
The reaction between barium chloride and sodium phosphate can be represented as:
\[ 3 \text{BaCl}_2 + 2 \text{Na}_3\text{PO}_4 \rightarrow 6 \text{NaCl} + \text{Ba}_3(\text{PO}_4)_2 \]

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

3. **Calculate the Required Moles of \( \text{Na}_3\text{PO}_4 \) for 0.5 moles of \( \text{BaCl}_2 \):**
To find out how much \( \text{Na}_3\text{PO}_4 \) is needed for 0.5 moles of \( \text{BaCl}_2 \):
\[ \text{From the ratio: } 3 \text{ moles of } \text{BaCl}_2 \text{ require } 2 \text{ moles of } \text{Na}_3\text{PO}_4 \]
Therefore, for 1 mole of \( \text{BaCl}_2 \):
\[ \text{Required } \text{Na}_3\text{PO}_4 = \frac{2}{3} \text{ moles} \]
For 0.5 moles of \( \text{BaCl}_2 \):
\[ \text{Required } \text{Na}_3\text{PO}_4 = \frac{2}{3} \times 0.5 = \frac{1}{3} \text{ moles} \approx 0.33 \text{ moles} \]

4. **Determine the Limiting Reagent:**
We have 0.2 moles of \( \text{Na}_3\text{PO}_4 \) available, which is less than the 0.33 moles required. Thus, \( \text{Na}_3\text{PO}_4 \) is the limiting reagent.

5. **Calculate the Maximum Moles of \( \text{Ba}_3(\text{PO}_4)_2 \) Formed:**
According to the balanced equation:
\[ 2 \text{ moles of } \text{Na}_3\text{PO}_4 \text{ produce } 1 \text{ mole of } \text{Ba}_3(\text{PO}_4)_2 \]
Therefore, for 0.2 moles of \( \text{Na}_3\text{PO}_4 \):
\[ \text{Moles of } \text{Ba}_3(\text{PO}_4)_2 = \frac{0.2}{2} = 0.1 \text{ moles} \]

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

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