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 \( \text{Ba}_3(\text{PO}_4)_2 \) that can be formed from the given reactants, we first need to write the balanced chemical equation for the reaction.

### Step 1: Write the balanced chemical equation
The balanced reaction between barium chloride (\( \text{BaCl}_2 \)) and sodium phosphate (\( \text{Na}_3\text{PO}_4 \)) is:

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

### Step 2: Identify the moles of reactants available
From the question, we have:
- 2 moles of \( \text{BaCl}_2 \)
- 1 mole of \( \text{Na}_3\text{PO}_4 \)

### Step 3: Determine the limiting reactant
To find out how many moles of \( \text{Ba}_3(\text{PO}_4)_2 \) can be formed, we need to determine which reactant is the limiting reactant.

From the balanced equation:
- 3 moles of \( \text{BaCl}_2 \) are needed to produce 1 mole of \( \text{Ba}_3(\text{PO}_4)_2 \).
- 2 moles of \( \text{Na}_3\text{PO}_4 \) are needed to produce 1 mole of \( \text{Ba}_3(\text{PO}_4)_2 \).

Now, we can calculate how many moles of \( \text{Ba}_3(\text{PO}_4)_2 \) can be produced from each reactant:

1. From \( \text{BaCl}_2 \):
\[ \text{Moles of } \text{Ba}_3(\text{PO}_4)_2 = \frac{2 \text{ moles } \text{BaCl}_2}{3} = \frac{2}{3} \text{ moles} \]

2. From \( \text{Na}_3\text{PO}_4 \):
\[ \text{Moles of } \text{Ba}_3(\text{PO}_4)_2 = \frac{1 \text{ mole } \text{Na}_3\text{PO}_4}{2} = \frac{1}{2} \text{ moles} \]

### Step 4: Compare the results
Now we compare the amounts calculated:
- From \( \text{BaCl}_2 \): \( \frac{2}{3} \approx 0.67 \) moles
- From \( \text{Na}_3\text{PO}_4 \): \( \frac{1}{2} = 0.5 \) moles

The limiting reactant is \( \text{Na}_3\text{PO}_4 \) because it produces fewer moles of \( \text{Ba}_3(\text{PO}_4)_2 \).

### Step 5: Conclusion
Thus, the maximum number of moles of \( \text{Ba}_3(\text{PO}_4)_2 \) that can be formed is:

\[ \text{Maximum moles of } \text{Ba}_3(\text{PO}_4)_2 = \frac{1}{2} \text{ moles} \]

### Final Answer
The maximum number of moles of \( \text{Ba}_3(\text{PO}_4)_2 \) that can be formed is **0.5 moles**.