If 0.50 mol of `BaCl_(2)` is mixed with 0.20 mol of `Na_(3)PO_(4)`, the maximum number of moles of `Ba_(3)(PO_(4))_(2)` that can be formed is

Options

A. `0.10`
B. `0.20`
C. `0.30`
D. `0.40`

Answer

To solve the problem, we need to determine the maximum number of moles of \( \text{Ba}_3(\text{PO}_4)_2 \) that can be formed when 0.50 moles of \( \text{BaCl}_2 \) is mixed with 0.20 moles of \( \text{Na}_3\text{PO}_4 \).

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction between \( \text{BaCl}_2 \) and \( \text{Na}_3\text{PO}_4 \) 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: Determine 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 \( \text{Ba}_3(\text{PO}_4)_2 \).

### Step 3: Identify the limiting reagent
We have:
- 0.50 moles of \( \text{BaCl}_2 \)
- 0.20 moles of \( \text{Na}_3\text{PO}_4 \)

Now, we need to find out how many moles of \( \text{Ba}_3(\text{PO}_4)_2 \) can be formed from each reactant.

**From \( \text{BaCl}_2 \):**
Using the stoichiometric ratio from the balanced equation:
\[ \text{Moles of } \text{Ba}_3(\text{PO}_4)_2 = \frac{0.50 \text{ moles of } \text{BaCl}_2}{3} = \frac{0.50}{3} \approx 0.167 \text{ moles} \]

**From \( \text{Na}_3\text{PO}_4 \):**
Using the stoichiometric ratio from the balanced equation:
\[ \text{Moles of } \text{Ba}_3(\text{PO}_4)_2 = \frac{0.20 \text{ moles of } \text{Na}_3\text{PO}_4}{2} = \frac{0.20}{2} = 0.10 \text{ moles} \]

### Step 4: Determine the limiting reagent
The limiting reagent is the one that produces the least amount of product. From our calculations:
- \( \text{BaCl}_2 \) can produce approximately 0.167 moles of \( \text{Ba}_3(\text{PO}_4)_2 \)
- \( \text{Na}_3\text{PO}_4 \) can produce 0.10 moles of \( \text{Ba}_3(\text{PO}_4)_2 \)

Since \( \text{Na}_3\text{PO}_4 \) produces the lesser amount, it is the limiting reagent.

### Step 5: Calculate the maximum moles of \( \text{Ba}_3(\text{PO}_4)_2 \)
Thus, the maximum number of moles of \( \text{Ba}_3(\text{PO}_4)_2 \) that can be formed is:
\[ \text{Maximum moles of } \text{Ba}_3(\text{PO}_4)_2 = 0.10 \text{ moles} \]

### 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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