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 step by step, we will first determine the amount of HCl that reacted with Zn and then calculate the weight of Zn that reacted with it.

### Step 1: Calculate the moles of NaOH used
Given:
- Normality of NaOH = \( \frac{N}{5} \)
- Volume of NaOH = 15 ml = 0.015 L

Using the formula for normality:
\[ \text{Moles of NaOH} = \text{Normality} \times \text{Volume in L} \]
\[ \text{Moles of NaOH} = \frac{1}{5} \times 0.015 = 0.003 \text{ moles} \]

### Step 2: Determine the moles of HCl neutralized by NaOH
The reaction between NaOH and HCl is a 1:1 reaction:
\[ \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]
Thus, the moles of HCl neutralized by NaOH is also 0.003 moles.

### Step 3: Calculate the total moles of HCl initially present
Given:
- Normality of HCl = \( \frac{N}{10} \)
- Volume of HCl = 40 ml = 0.040 L

Using the formula for normality:
\[ \text{Moles of HCl} = \text{Normality} \times \text{Volume in L} \]
\[ \text{Moles of HCl} = \frac{1}{10} \times 0.040 = 0.004 \text{ moles} \]

### Step 4: Calculate the moles of HCl that reacted with Zn
The moles of HCl that reacted with Zn can be calculated by subtracting the moles of HCl that were neutralized by NaOH from the total moles of HCl:
\[ \text{Moles of HCl that reacted with Zn} = \text{Total moles of HCl} - \text{Moles of HCl neutralized} \]
\[ \text{Moles of HCl that reacted with Zn} = 0.004 - 0.003 = 0.001 \text{ moles} \]

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

### Step 6: Calculate the weight of Zn that reacted
Using the molar mass of Zn (approximately 65.38 g/mol):
\[ \text{Weight of Zn} = \text{Moles of Zn} \times \text{Molar mass of Zn} \]
\[ \text{Weight of Zn} = 0.0005 \times 65.38 \approx 0.03269 \text{ grams} \]

### Final Answer
The weight of Zn that reacts with HCl is approximately **0.03269 grams**.