According to following reaction:
`A + BO_(3) rArr A_(3)O_(4) + B_(2)O_(3)`
The number of moles of moles of `A_(3)O_(4)` produced if 1 mole of A is missed with mole of `BO_(3)` is:

Options

A. 3
B. 1
C. `1/3`
D. `2/3`

Answer

To determine the number of moles of \( A_3O_4 \) produced when 1 mole of \( A \) reacts with 1 mole of \( BO_3 \), we first need to balance the given chemical equation:

### Step 1: Write the unbalanced equation
The unbalanced equation is:
\[ A + BO_3 \rightarrow A_3O_4 + B_2O_3 \]

### Step 2: Balance the equation
To balance the equation, we need to ensure that the number of atoms of each element is the same on both sides.

1. **Balancing A**: We have 1 atom of \( A \) on the left and 3 atoms of \( A \) in \( A_3O_4 \) on the right. Therefore, we need to put a coefficient of 3 in front of \( A \) on the left:
\[ 3A + BO_3 \rightarrow A_3O_4 + B_2O_3 \]

2. **Balancing B**: We have 1 atom of \( B \) in \( BO_3 \) on the left and 2 atoms of \( B \) in \( B_2O_3 \) on the right. Therefore, we need to put a coefficient of 2 in front of \( BO_3 \):
\[ 3A + 2BO_3 \rightarrow A_3O_4 + B_2O_3 \]

3. **Balancing O**: Now, we count the oxygen atoms. We have 6 oxygen atoms from \( 2BO_3 \) on the left and 4 from \( A_3O_4 \) and 3 from \( B_2O_3 \) on the right, totaling 7. We need to adjust our coefficients to balance the oxygen as well.

After balancing, the final balanced equation is:
\[ 3A + 2BO_3 \rightarrow A_3O_4 + 2B_2O_3 \]

### Step 3: Identify the limiting reagent
Now, we need to find out how many moles of \( A_3O_4 \) can be produced from the available reactants.

From the balanced equation:
- 3 moles of \( A \) react with 2 moles of \( BO_3 \) to produce 1 mole of \( A_3O_4 \).

Given that we have:
- 1 mole of \( A \)
- 1 mole of \( BO_3 \)

We can find out how many moles of \( A_3O_4 \) can be produced based on the limiting reagent.

1. **From \( A \)**:
- If 3 moles of \( A \) produce 1 mole of \( A_3O_4 \), then 1 mole of \( A \) would produce:
\[ \frac{1 \text{ mole of } A}{3} = \frac{1}{3} \text{ moles of } A_3O_4 \]

2. **From \( BO_3 \)**:
- If 2 moles of \( BO_3 \) produce 1 mole of \( A_3O_4 \), then 1 mole of \( BO_3 \) would produce:
\[ \frac{1 \text{ mole of } BO_3}{2} = \frac{1}{2} \text{ moles of } A_3O_4 \]

### Step 4: Determine the limiting reagent
Now, we compare the amounts produced:
- From 1 mole of \( A \), we can produce \( \frac{1}{3} \) moles of \( A_3O_4 \).
- From 1 mole of \( BO_3 \), we can produce \( \frac{1}{2} \) moles of \( A_3O_4 \).

Since \( \frac{1}{3} \) is less than \( \frac{1}{2} \), \( A \) is the limiting reagent.

### Step 5: Conclusion
Thus, the number of moles of \( A_3O_4 \) produced when 1 mole of \( A \) reacts with 1 mole of \( BO_3 \) is:
\[ \frac{1}{3} \text{ moles of } A_3O_4 \]