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

Options

A. `0.7`
B. `0.5`
C. `0.30`
D. `0.1`

Answer

To solve the problem, we need to determine the maximum number of moles of barium phosphate, \( \text{Ba}_3(\text{PO}_4)_2 \), that can be formed when mixing 0.5 moles of barium chloride, \( \text{BaCl}_2 \), 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 written as:
\[ 3 \text{BaCl}_2 + 2 \text{Na}_3\text{PO}_4 \rightarrow \text{Ba}_3(\text{PO}_4)_2 + 6 \text{NaCl} \]
This equation indicates that 3 moles of \( \text{BaCl}_2 \) react with 2 moles of \( \text{Na}_3\text{PO}_4 \) to produce 1 mole of \( \text{Ba}_3(\text{PO}_4)_2 \).

2. **Identify the Limiting Reagent:**
- From the balanced equation, we see that:
- 3 moles of \( \text{BaCl}_2 \) are needed for every 2 moles of \( \text{Na}_3\text{PO}_4 \).
- We have:
- 0.5 moles of \( \text{BaCl}_2 \)
- 0.2 moles of \( \text{Na}_3\text{PO}_4 \)

To find out which reactant is limiting, we can calculate how much of each reactant is required:
- For 0.2 moles of \( \text{Na}_3\text{PO}_4 \):
- Using the ratio from the balanced equation, we need:
\[ \text{BaCl}_2 \text{ required} = \frac{3}{2} \times 0.2 = 0.3 \text{ moles} \]
- Since we have 0.5 moles of \( \text{BaCl}_2 \), it is in excess. Therefore, \( \text{Na}_3\text{PO}_4 \) is the limiting reagent.

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

### 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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