A piece of Zn is dissolved in 40 ml of `(N)/(10)HCl` completely. The excess of acid was neutralized by 15 ml of `(N)/(5)NaOH`. The weight of Zn which react with `HCl` is :

Options

A. 65 gm
B. 0.065 gm
C. 0.0325 gm
D. 32.25 gm

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the moles of NaOH used
Given:
- Normality of NaOH = \( \frac{N}{5} \)
- Volume of NaOH = 15 ml

First, we convert the volume from ml to liters:
\[ \text{Volume of NaOH in liters} = \frac{15}{1000} = 0.015 \text{ L} \]

Now, we calculate the molarity of NaOH:
\[ \text{Molarity of NaOH} = \text{Normality} \times \text{n-factor} = \frac{1}{5} \times 1 = 0.2 \text{ M} \]

Next, we calculate the moles of NaOH:
\[ \text{Moles of NaOH} = \text{Molarity} \times \text{Volume in L} = 0.2 \times 0.015 = 0.003 \text{ moles} \]

### Step 2: Calculate the moles of HCl neutralized
From the reaction:
\[ \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]
1 mole of NaOH neutralizes 1 mole of HCl. Therefore, the moles of HCl neutralized by NaOH is also:
\[ \text{Moles of HCl neutralized} = 0.003 \text{ moles} \]

### Step 3: Calculate the total moles of HCl initially present
Given:
- Normality of HCl = \( \frac{N}{10} \)
- Volume of HCl = 40 ml

Convert the volume from ml to liters:
\[ \text{Volume of HCl in liters} = \frac{40}{1000} = 0.040 \text{ L} \]

Calculate the molarity of HCl:
\[ \text{Molarity of HCl} = \frac{1}{10} \times 1 = 0.1 \text{ M} \]

Now, calculate the total moles of HCl:
\[ \text{Moles of HCl} = \text{Molarity} \times \text{Volume in L} = 0.1 \times 0.040 = 0.004 \text{ moles} \]

### Step 4: Calculate the excess moles of HCl
The excess moles of HCl can be calculated as follows:
\[ \text{Excess moles of HCl} = \text{Total moles of HCl} - \text{Moles of HCl neutralized} \]
\[ \text{Excess moles of HCl} = 0.004 - 0.003 = 0.001 \text{ moles} \]

### Step 5: Calculate the moles of Zn that reacted with HCl
The reaction between Zn and HCl is:
\[ \text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2 \]
From the stoichiometry, 1 mole of Zn reacts with 2 moles of HCl. Thus, the moles of Zn that reacted can be calculated as:
\[ \text{Moles of Zn} = \frac{\text{Excess moles of HCl}}{2} = \frac{0.001}{2} = 0.0005 \text{ moles} \]

### Step 6: Calculate the weight of Zn that reacted
The atomic weight of Zn is approximately 65.38 g/mol. Therefore, the weight of Zn that reacted is:
\[ \text{Weight of Zn} = \text{Moles of Zn} \times \text{Atomic weight of Zn} = 0.0005 \times 65.38 = 0.03269 \text{ g} \]

### Final Answer
The weight of Zn which reacts with HCl is approximately:
\[ \text{Weight of Zn} \approx 0.0325 \text{ g} \]