The maximum number of moles of `Ba_(3)(PO_4)_(2)` that can be formed if 2 mole `BaCl_2` is mixed with 1 mole `Na_3PO_(4)` is

Options

A. 0.66
B. 0.25
C. 0.33
D. 0.5

Answer

To determine the maximum number of moles of Barium Phosphate (Ba₃(PO₄)₂) that can be formed when 2 moles of BaCl₂ are mixed with 1 mole of Na₃PO₄, we will follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between barium chloride (BaCl₂) and sodium phosphate (Na₃PO₄) can be represented as:
\[ 3 \text{BaCl}_2 + 2 \text{Na}_3\text{PO}_4 \rightarrow \text{Ba}_3(\text{PO}_4)_2 + 6 \text{NaCl} \]

### Step 2: Identify the limiting reagent
From the balanced equation, we can see that:
- 3 moles of BaCl₂ are required to produce 1 mole of Ba₃(PO₄)₂.
- 2 moles of Na₃PO₄ are required to produce 1 mole of Ba₃(PO₄)₂.

Given that we have 2 moles of BaCl₂ and 1 mole of Na₃PO₄, we need to determine which reactant will limit the formation of Ba₃(PO₄)₂.

### Step 3: Calculate the moles of Ba₃(PO₄)₂ produced from BaCl₂
Using the stoichiometry from the balanced equation:
- From 3 moles of BaCl₂, we can produce 1 mole of Ba₃(PO₄)₂.
- Therefore, from 2 moles of BaCl₂, the amount of Ba₃(PO₄)₂ produced can be calculated as follows:

\[ \text{Moles of Ba}_3(\text{PO}_4)_2 = \left(\frac{1 \text{ mole Ba}_3(\text{PO}_4)_2}{3 \text{ moles BaCl}_2}\right) \times 2 \text{ moles BaCl}_2 = \frac{2}{3} \text{ moles Ba}_3(\text{PO}_4)_2 \]

### Step 4: Calculate the moles of Ba₃(PO₄)₂ produced from Na₃PO₄
Now, from the balanced equation:
- 2 moles of Na₃PO₄ are required to produce 1 mole of Ba₃(PO₄)₂.
- Since we have 1 mole of Na₃PO₄, we can find the moles of Ba₃(PO₄)₂ produced:

\[ \text{Moles of Ba}_3(\text{PO}_4)_2 = \left(\frac{1 \text{ mole Ba}_3(\text{PO}_4)_2}{2 \text{ moles Na}_3\text{PO}_4}\right) \times 1 \text{ mole Na}_3\text{PO}_4 = \frac{1}{2} \text{ moles Ba}_3(\text{PO}_4)_2 \]

### Step 5: Determine the maximum moles of Ba₃(PO₄)₂
Now we compare the amounts produced from both reactants:
- From BaCl₂: \( \frac{2}{3} \) moles of Ba₃(PO₄)₂
- From Na₃PO₄: \( \frac{1}{2} \) moles of Ba₃(PO₄)₂

To find the limiting reagent, we can convert \( \frac{1}{2} \) to a fraction with a common denominator:
- \( \frac{1}{2} = \frac{3}{6} \)
- \( \frac{2}{3} = \frac{4}{6} \)

Since \( \frac{1}{2} \) (or \( \frac{3}{6} \)) is less than \( \frac{2}{3} \) (or \( \frac{4}{6} \)), Na₃PO₄ is the limiting reagent.

### Conclusion
The maximum number of moles of Ba₃(PO₄)₂ that can be formed is \( \frac{1}{2} \) moles.

### Final Answer
The maximum number of moles of Ba₃(PO₄)₂ that can be formed is **0.5 moles**.