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

Options

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

Explanation

To determine the maximum number of moles of \( \text{Ba}_3(\text{PO}_4)_2 \) that can be formed from the reaction between \( \text{BaCl}_2 \) and \( \text{Na}_3\text{PO}_4 \), we will follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between barium chloride and sodium phosphate can be represented as follows:

\[ 3 \text{BaCl}_2 + 2 \text{Na}_3\text{PO}_4 \rightarrow \text{Ba}_3(\text{PO}_4)_2 + 6 \text{NaCl} \]

### Step 2: Identify the mole ratios from the balanced equation
From the balanced equation, we can see the following mole ratios:
- 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: Determine the limiting reagent
We have:
- 0.5 moles of \( \text{BaCl}_2 \)
- 0.20 moles of \( \text{Na}_3\text{PO}_4 \)

Now, we need to calculate how much of each reactant is required to produce \( \text{Ba}_3(\text{PO}_4)_2 \).

#### For \( \text{Na}_3\text{PO}_4 \):
From the balanced equation, 2 moles of \( \text{Na}_3\text{PO}_4 \) are required for every 1 mole of \( \text{Ba}_3(\text{PO}_4)_2 \). Thus, for 0.20 moles of \( \text{Na}_3\text{PO}_4 \):
- Required \( \text{BaCl}_2 \) = \( \frac{3}{2} \times 0.20 = 0.30 \) moles of \( \text{BaCl}_2 \)

#### For \( \text{BaCl}_2 \):
We have 0.5 moles of \( \text{BaCl}_2 \), which is more than the 0.30 moles required for the available \( \text{Na}_3\text{PO}_4 \).

### Conclusion on the limiting reagent:
Since \( \text{Na}_3\text{PO}_4 \) is the limiting reagent, we will use it to calculate the maximum moles of \( \text{Ba}_3(\text{PO}_4)_2 \) that can be formed.

### Step 4: Calculate the maximum moles of \( \text{Ba}_3(\text{PO}_4)_2 \)
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, 0.20 moles of \( \text{Na}_3\text{PO}_4 \) will produce:
\[ \text{Moles of } \text{Ba}_3(\text{PO}_4)_2 = \frac{0.20}{2} = 0.10 \text{ moles} \]

### Final Answer:
The maximum number of moles of \( \text{Ba}_3(\text{PO}_4)_2 \) that can be formed is **0.10 moles**.

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