Which of the following statements is/are correct

Options

A. The reaction.
`Bi + 4HNO_(3) + 3H_(2) O rarr Bi (NO_(3))_(3). 5H_(2) O + NO`
`2.09 g` of `Bi` in `HNO_(3)` produces `48.5g` of bismuth nitrate.
(Atomic weight `Bi = 209 g, Mw "of" Bi (NO_(3))_(3).5H_(2)O = 485 g)`
B. `4.0 g` of 63% `HNO_(3)` by mass is required to react with `2.09 g` of `Bi`
C. The volume of `NO` gas produced at `STP` (1 bar `273 K`) is `0.227 L`
D. The volume of `NO` gas produced at `STP` (1 bar, `298K`) is `0.247 L`.

Explanation

To determine which of the statements regarding the reaction \( \text{Bi} + 4 \text{HNO}_3 + 3 \text{H}_2\text{O} \rightarrow \text{Bi(NO}_3)_3 \cdot 5 \text{H}_2\text{O} + \text{NO} \) are correct, we will analyze each statement step by step.

### Step 1: Analyze the First Statement
The first statement claims that 48.5 grams of bismuth nitrate is produced by 20.9 grams of bismuth.

1. **Molar Mass Calculation**:
- Molar mass of bismuth (Bi) = 209 g/mol.
- Molar mass of bismuth nitrate (Bi(NO₃)₃·5H₂O) = 485 g/mol.

2. **Stoichiometry**:
- From the balanced equation, 1 mole of Bi produces 1 mole of Bi(NO₃)₃·5H₂O.
- Therefore, 209 g of Bi produces 485 g of Bi(NO₃)₃·5H₂O.

3. **Calculating the Mass of Bismuth for 48.5 g of Bismuth Nitrate**:
- To find how much Bi is needed to produce 48.5 g of Bi(NO₃)₃·5H₂O:
\[ \text{Mass of Bi} = \frac{209 \text{ g}}{485 \text{ g}} \times 48.5 \text{ g} = 20.9 \text{ g} \]
- This means that 20.9 g of Bi produces 48.5 g of Bi(NO₃)₃·5H₂O.

**Conclusion for Statement 1**: The first statement is **incorrect**.

### Step 2: Analyze the Second Statement
The second statement claims that 4 grams of 63% HNO₃ by mass is required to react with 2.09 grams of bismuth.

1. **Molar Mass of HNO₃**:
- Molar mass of HNO₃ = 63 g/mol.

2. **Stoichiometry**:
- From the balanced equation, 1 mole of Bi reacts with 4 moles of HNO₃.
- Therefore, 209 g of Bi requires:
\[ 4 \times 63 \text{ g} = 252 \text{ g of HNO}_3 \]

3. **Calculating HNO₃ Required for 2.09 g of Bismuth**:
- For 2.09 g of Bi:
\[ \text{HNO}_3 \text{ required} = \frac{252 \text{ g}}{209 \text{ g}} \times 2.09 \text{ g} = 2.52 \text{ g} \]
- Since 63% HNO₃ means that in 100 g of solution, there are 63 g of HNO₃:
\[ \text{Mass of HNO}_3 = 4 \text{ g of 63% solution} \Rightarrow \text{HNO}_3 = 4 \times \frac{63}{100} = 2.52 \text{ g} \]

**Conclusion for Statement 2**: The second statement is **correct**.

### Step 3: Analyze the Third Statement
The third statement claims that the volume of NO gas produced at STP is 0.227 liters.

1. **Moles of NO Produced**:
- From the balanced equation, 1 mole of Bi produces 1 mole of NO.
- Therefore, 209 g of Bi produces 1 mole of NO.

2. **Volume of NO at STP**:
- At STP, 1 mole of gas occupies 22.4 liters.
- Thus, the volume of NO produced is:
\[ \text{Volume of NO} = 1 \text{ mole} \times 22.4 \text{ L/mole} = 22.4 \text{ L} \]

**Conclusion for Statement 3**: The third statement is **incorrect**.

### Final Conclusion
The only correct statement is the second one. Therefore, the correct option is **option 2**.