`5.00` moles of hydrogen gas, 3 moles of white phosphorus `{P_(4)(s)}` and 12 moles of oxygen gas are taken in a sealed flask and allowed to react as follows :
`H_(2)(g)+P_(4)(s)+O_(2)(g)rarrH_(2)PO_(4)`
Determine the moles of ortho-phosphoric acid that can be produced, considering that the reaction occurs in `90 %` yield.

Answer

To determine the moles of ortho-phosphoric acid (H₃PO₄) that can be produced from the given reactants, we will follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction is:
\[ H_2(g) + P_4(s) + 2O_2(g) \rightarrow 2H_3PO_4(aq) \]

### Step 2: Identify the moles of each reactant
From the problem, we have:
- Moles of \( H_2 \) = 5.00 moles
- Moles of \( P_4 \) = 3.00 moles
- Moles of \( O_2 \) = 12.00 moles

### Step 3: Determine the stoichiometric ratios
From the balanced equation, we can see the stoichiometric ratios:
- 1 mole of \( H_2 \) reacts with 1 mole of \( P_4 \) and 2 moles of \( O_2 \) to produce 2 moles of \( H_3PO_4 \).

### Step 4: Calculate the limiting reactant
Now, we will calculate how many moles of \( H_3PO_4 \) can be produced based on each reactant:

1. **From \( H_2 \)**:
- Moles of \( H_3PO_4 \) produced = \( 5.00 \, \text{moles of } H_2 \times \frac{2 \, \text{moles of } H_3PO_4}{1 \, \text{mole of } H_2} = 10.00 \, \text{moles of } H_3PO_4 \)

2. **From \( P_4 \)**:
- Moles of \( H_3PO_4 \) produced = \( 3.00 \, \text{moles of } P_4 \times \frac{2 \, \text{moles of } H_3PO_4}{1 \, \text{mole of } P_4} = 6.00 \, \text{moles of } H_3PO_4 \)

3. **From \( O_2 \)**:
- Moles of \( H_3PO_4 \) produced = \( 12.00 \, \text{moles of } O_2 \times \frac{2 \, \text{moles of } H_3PO_4}{2 \, \text{moles of } O_2} = 12.00 \, \text{moles of } H_3PO_4 \)

### Step 5: Identify the limiting reactant
The limiting reactant is the one that produces the least amount of product:
- \( H_2 \) produces 10.00 moles
- \( P_4 \) produces 6.00 moles (limiting reactant)
- \( O_2 \) produces 12.00 moles

Thus, \( P_4 \) is the limiting reactant.

### Step 6: Calculate the theoretical yield of \( H_3PO_4 \)
The theoretical yield of \( H_3PO_4 \) based on the limiting reactant \( P_4 \) is 6.00 moles.

### Step 7: Adjust for the percentage yield
Since the reaction occurs at a 90% yield:
\[ \text{Actual yield of } H_3PO_4 = \text{Theoretical yield} \times \text{Percentage yield} \]
\[ \text{Actual yield of } H_3PO_4 = 6.00 \, \text{moles} \times 0.90 = 5.40 \, \text{moles} \]

### Final Answer
The moles of ortho-phosphoric acid (H₃PO₄) that can be produced is **5.40 moles**.

---