2 g of brass containing Cu and Zn only reacts with 3M `HNO_(3)` solution. Following are the reactions taking place `Cu(s) + HNO_(3) (aq) to Cu^(2+) (aq) +NO_(2)(g) + H_(2)O(I)`
`Zn(s) + H^(+)(aq) + NO_(3)^(-)(aq) to NH_(4)^(+) + Zn^(2+)(aq) + H_(2)O(l)`
The liberated `NO_(2)(g)` was found to be 1.04 L at `25^(@)`C and 1 atm `[Cu=63.5 , Zn=65.4]`
How many grams of `NH_(4)NO_(3)` will be obtained in the above reaction ?

Options

A. 0.405 g
B. 0.0428 g
C. 0.2018 g
D. 0.358 g

Answer

To solve the problem, we need to follow these steps:

### Step 1: Calculate the moles of NO₂ produced
We know that the volume of NO₂ gas produced is 1.04 L at 25°C and 1 atm. We can use the ideal gas equation to find the number of moles of NO₂.

**Formula:**
\[ PV = nRT \]

Where:
- \( P = 1 \, \text{atm} \)
- \( V = 1.04 \, \text{L} \)
- \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \)
- \( T = 25°C = 298 \, \text{K} \)

**Calculation:**
\[ n = \frac{PV}{RT} \]
\[ n = \frac{(1)(1.04)}{(0.0821)(298)} \]
\[ n = \frac{1.04}{24.4758} \]
\[ n \approx 0.0425 \, \text{mol} \]

### Step 2: Relate moles of NO₂ to moles of Cu
From the reaction, we see that:
\[ \text{Cu} + 4 \text{HNO}_3 \rightarrow \text{Cu(NO}_3\text{)}_2 + 2 \text{NO}_2 + 2 \text{H}_2\text{O} \]

From the stoichiometry of the reaction, 1 mole of Cu produces 2 moles of NO₂. Therefore, the moles of Cu can be calculated as follows:
\[ \text{Moles of Cu} = \frac{1}{2} \times \text{Moles of NO}_2 \]
\[ \text{Moles of Cu} = \frac{1}{2} \times 0.0425 \]
\[ \text{Moles of Cu} \approx 0.02125 \, \text{mol} \]

### Step 3: Calculate the mass of Cu
Using the molar mass of Cu (63.5 g/mol):
\[ \text{Mass of Cu} = \text{Moles of Cu} \times \text{Molar mass of Cu} \]
\[ \text{Mass of Cu} = 0.02125 \times 63.5 \]
\[ \text{Mass of Cu} \approx 1.3496 \, \text{g} \]

### Step 4: Calculate the mass of Zn
The total mass of brass is given as 2 g. Therefore, the mass of Zn can be calculated as:
\[ \text{Mass of Zn} = \text{Total mass} - \text{Mass of Cu} \]
\[ \text{Mass of Zn} = 2 - 1.3496 \]
\[ \text{Mass of Zn} \approx 0.6504 \, \text{g} \]

### Step 5: Calculate the moles of Zn
Using the molar mass of Zn (65.4 g/mol):
\[ \text{Moles of Zn} = \frac{\text{Mass of Zn}}{\text{Molar mass of Zn}} \]
\[ \text{Moles of Zn} = \frac{0.6504}{65.4} \]
\[ \text{Moles of Zn} \approx 0.00995 \, \text{mol} \]

### Step 6: Relate moles of Zn to moles of NH₄NO₃
From the reaction:
\[ \text{Zn} + 4 \text{HNO}_3 \rightarrow \text{NH}_4\text{NO}_3 + \text{Zn(NO}_3\text{)}_2 + 3 \text{H}_2\text{O} \]

From the stoichiometry, 1 mole of Zn produces 1 mole of NH₄NO₃. Therefore:
\[ \text{Moles of NH}_4\text{NO}_3 = \text{Moles of Zn} \]
\[ \text{Moles of NH}_4\text{NO}_3 \approx 0.00995 \, \text{mol} \]

### Step 7: Calculate the mass of NH₄NO₃
Using the molar mass of NH₄NO₃ (80 g/mol):
\[ \text{Mass of NH}_4\text{NO}_3 = \text{Moles of NH}_4\text{NO}_3 \times \text{Molar mass of NH}_4\text{NO}_3 \]
\[ \text{Mass of NH}_4\text{NO}_3 = 0.00995 \times 80 \]
\[ \text{Mass of NH}_4\text{NO}_3 \approx 0.796 \, \text{g} \]

### Final Result
The mass of NH₄NO₃ obtained from the reaction is approximately **0.796 g**.