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. 0.7
B. 0.5
C. 0.2
D. 0.1

Answer

To solve the problem of how many moles of \( Ba_3(PO_4)_2 \) can be formed from the reaction of \( BaCl_2 \) and \( Na_3PO_4 \), we will follow these steps:

### Step 1: Write the balanced chemical equation

The reaction between barium chloride and sodium phosphate can be represented as:
\[ 3 BaCl_2 + 2 Na_3PO_4 \rightarrow Ba_3(PO_4)_2 + 6 NaCl \]

### Step 2: Identify the moles of reactants

From the problem, we have:
- Moles of \( BaCl_2 = 0.50 \, \text{mol} \)
- Moles of \( Na_3PO_4 = 0.20 \, \text{mol} \)

### Step 3: Determine the stoichiometric ratios

From the balanced equation, we see that:
- 3 moles of \( BaCl_2 \) react with 2 moles of \( Na_3PO_4 \) to produce 1 mole of \( Ba_3(PO_4)_2 \).

### Step 4: Calculate the limiting reactant

To find the limiting reactant, we need to calculate how many moles of \( Ba_3(PO_4)_2 \) can be produced from each reactant.

1. **From \( BaCl_2 \)**:
\[ \text{Moles of } Ba_3(PO_4)_2 \text{ from } BaCl_2 = \frac{0.50 \, \text{mol} \, BaCl_2}{3} = 0.1667 \, \text{mol} \]

2. **From \( Na_3PO_4 \)**:
\[ \text{Moles of } Ba_3(PO_4)_2 \text{ from } Na_3PO_4 = \frac{0.20 \, \text{mol} \, Na_3PO_4}{2} = 0.10 \, \text{mol} \]

### Step 5: Identify the limiting reactant

Since \( Na_3PO_4 \) produces fewer moles of \( Ba_3(PO_4)_2 \) (0.10 mol), it is the limiting reactant.

### Step 6: Calculate the maximum number of moles of \( Ba_3(PO_4)_2 \)

The maximum number of moles of \( Ba_3(PO_4)_2 \) that can be formed is determined by the limiting reactant:
\[ \text{Maximum moles of } Ba_3(PO_4)_2 = 0.10 \, \text{mol} \]

### Final Answer

The maximum number of moles of \( Ba_3(PO_4)_2 \) that can be formed is **0.10 mol**.

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