446 g of PbO , 46 g of `NO_(2)` and 16 g of `O_(2)` are allowed to react according to the equation :-
`PbO+2NO_(2)+(1)/(2)O_(2)toPb(NO_(3))_(2)`
The amount of `Pb(NO_(3))_(2)` that can be produced is (At . Wt. of Pb =207) :-

Options

A. (a)331 g
B. (b)662 g
C. (c)165.5 g
D. (d)none of these

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the moles of each reactant.

1. **For PbO:**
- Given mass = 446 g
- Molar mass of PbO = Atomic mass of Pb + Atomic mass of O = 207 + 16 = 223 g/mol
- Moles of PbO = Given mass / Molar mass = 446 g / 223 g/mol = 2 moles

2. **For NO2:**
- Given mass = 46 g
- Molar mass of NO2 = Atomic mass of N + 2 * Atomic mass of O = 14 + 2*16 = 46 g/mol
- Moles of NO2 = Given mass / Molar mass = 46 g / 46 g/mol = 1 mole

3. **For O2:**
- Given mass = 16 g
- Molar mass of O2 = 2 * Atomic mass of O = 2*16 = 32 g/mol
- Moles of O2 = Given mass / Molar mass = 16 g / 32 g/mol = 0.5 moles

### Step 2: Identify the limiting reagent.

The balanced chemical equation is:
\[ \text{PbO} + 2 \text{NO}_2 + \frac{1}{2} \text{O}_2 \rightarrow \text{Pb(NO}_3\text{)}_2 \]

From the equation:
- 1 mole of PbO reacts with 2 moles of NO2 and 0.5 moles of O2.

Now, we compare the available moles with the stoichiometric coefficients:
- For PbO: 2 moles available, requires 1 mole (sufficient)
- For NO2: 1 mole available, requires 2 moles (limiting)
- For O2: 0.5 moles available, requires 0.5 moles (sufficient)

Since NO2 is the limiting reagent, we will use it to calculate the amount of Pb(NO3)2 produced.

### Step 3: Calculate the amount of Pb(NO3)2 produced.

According to the reaction:
- 2 moles of NO2 produce 1 mole of Pb(NO3)2.

Since we have 1 mole of NO2:
- Moles of Pb(NO3)2 produced = 1 mole NO2 * (1 mole Pb(NO3)2 / 2 moles NO2) = 0.5 moles Pb(NO3)2

### Step 4: Calculate the mass of Pb(NO3)2 produced.

1. **Molar mass of Pb(NO3)2:**
- Molar mass = Atomic mass of Pb + 2 * (Atomic mass of N + 3 * Atomic mass of O)
- Molar mass = 207 + 2 * (14 + 3*16) = 207 + 2 * (14 + 48) = 207 + 2 * 62 = 207 + 124 = 331 g/mol

2. **Mass of Pb(NO3)2 produced:**
- Mass = Moles * Molar mass = 0.5 moles * 331 g/mol = 165.5 g

### Final Answer:
The amount of Pb(NO3)2 that can be produced is **165.5 g**.

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