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]`
The percentage by mass of Cu in brass was

Options

A. `67%`
B. `70%`
C. `80%`
D. `90%`

Answer

To solve the problem of determining the percentage by mass of copper (Cu) in a 2 g sample of brass (which contains only copper and zinc), we can follow these steps:

### Step 1: Understand the reactions
The reactions taking place when brass reacts with nitric acid (HNO3) are:
1. For copper:
\[ Cu(s) + 4HNO_3(aq) \rightarrow Cu(NO_3)_2(aq) + 2NO_2(g) + 2H_2O(l) \]
2. For zinc:
\[ Zn(s) + 2H^+(aq) + 2NO_3^-(aq) \rightarrow NH_4^+(aq) + Zn^{2+}(aq) + 2H_2O(l) \]

### Step 2: Calculate moles of NO2 produced
Given that 1.04 L of NO2 gas is liberated at 25°C and 1 atm, we can use the ideal gas law to find the number of moles of NO2 produced.

Using the formula:
\[ PV = nRT \]
Where:
- \( P = 1 \, \text{atm} \)
- \( V = 1.04 \, \text{L} \)
- \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \)
- \( T = 25°C = 298 \, \text{K} \)

Rearranging for \( n \):
\[ n = \frac{PV}{RT} \]
Substituting the values:
\[ n = \frac{(1 \, \text{atm})(1.04 \, \text{L})}{(0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1})(298 \, \text{K})} \]
Calculating this gives:
\[ n = \frac{1.04}{24.4758} \approx 0.0425 \, \text{mol} \]

### Step 3: Relate moles of NO2 to moles of Cu
From the balanced reaction for copper, we see that 1 mole of Cu produces 2 moles of NO2. Therefore, the moles of Cu can be calculated as:
\[ \text{Moles of Cu} = \frac{1}{2} \times \text{Moles of NO2} = \frac{1}{2} \times 0.0425 \approx 0.02125 \, \text{mol} \]

### Step 4: Calculate the mass of Cu
Using the molar mass of Cu (63.5 g/mol):
\[ \text{Mass of Cu} = \text{Moles of Cu} \times \text{Molar mass of Cu} = 0.02125 \, \text{mol} \times 63.5 \, \text{g/mol} \approx 1.3496 \, \text{g} \]

### Step 5: Calculate the percentage by mass of Cu in brass
The percentage by mass of Cu in the brass sample can be calculated as:
\[ \text{Percentage of Cu} = \left( \frac{\text{Mass of Cu}}{\text{Total mass of brass}} \right) \times 100 = \left( \frac{1.3496 \, \text{g}}{2 \, \text{g}} \right) \times 100 \approx 67.48\% \]

### Final Answer
The percentage by mass of Cu in brass is approximately **67.4%**.

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