A solution of HCI is prepared by dissolving 7.30 g of hydrogen chloride gas in 100 ml of water. Find the molarity of the solution. If `50.0 cm^3` of this solution is treated with 3.50 g of zinc, what volume of `H_2` measured at S.T.P. will be evolved ?

Answer

To solve the problem step by step, we will first calculate the molarity of the HCl solution and then determine the volume of hydrogen gas evolved when zinc reacts with the HCl solution.

### Step 1: Calculate the Molarity of the HCl Solution

**Formula for Molarity (M):**
\[ M = \frac{W}{M_w \times V} \]
Where:
- \( W \) = weight of solute (in grams)
- \( M_w \) = molar mass of the solute (in g/mol)
- \( V \) = volume of solution (in liters)

**Given:**
- Weight of HCl, \( W = 7.30 \, \text{g} \)
- Volume of solution, \( V = 100 \, \text{ml} = 0.1 \, \text{L} \)
- Molar mass of HCl, \( M_w = 1 + 35.5 = 36.5 \, \text{g/mol} \)

**Substituting the values:**
\[ M = \frac{7.30}{36.5 \times 0.1} = \frac{7.30}{3.65} = 2 \, \text{M} \]

### Step 2: Calculate the Amount of HCl in 50 ml of the Solution

**Using the molarity to find the weight of HCl in 50 ml:**
\[ \text{Weight of HCl} = M \times M_w \times V \]
Where \( V = 50 \, \text{ml} = 0.050 \, \text{L} \).

**Substituting the values:**
\[ \text{Weight of HCl} = 2 \, \text{mol/L} \times 36.5 \, \text{g/mol} \times 0.050 \, \text{L} = 3.65 \, \text{g} \]

### Step 3: Determine the Limiting Reagent

**Reaction Equation:**
\[ \text{Zn} + 2 \text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2 \]

**Molar Mass of Zinc:**
- Molar mass of Zn = 65.5 g/mol

**Calculate moles of Zinc:**
\[ \text{Moles of Zn} = \frac{3.50 \, \text{g}}{65.5 \, \text{g/mol}} \approx 0.0535 \, \text{mol} \]

**From the reaction, 1 mole of Zn reacts with 2 moles of HCl. Thus, moles of HCl needed:**
\[ \text{Moles of HCl needed} = 2 \times 0.0535 \approx 0.107 \, \text{mol} \]

**Now, calculate the moles of HCl available:**
\[ \text{Moles of HCl in 3.65 g} = \frac{3.65 \, \text{g}}{36.5 \, \text{g/mol}} \approx 0.100 \, \text{mol} \]

**Comparing the moles:**
- Moles of HCl needed = 0.107 mol
- Moles of HCl available = 0.100 mol

Since we have less HCl than needed, **HCl is the limiting reagent**.

### Step 4: Calculate the Volume of Hydrogen Gas Produced

**Using the stoichiometry of the reaction:**
- 2 moles of HCl produce 1 mole of H2.
- From 0.100 moles of HCl, the moles of H2 produced:
\[ \text{Moles of H}_2 = \frac{0.100}{2} = 0.050 \, \text{mol} \]

**At STP, 1 mole of gas occupies 22.4 L. Therefore, the volume of H2 produced is:**
\[ \text{Volume of H}_2 = 0.050 \, \text{mol} \times 22.4 \, \text{L/mol} = 1.12 \, \text{L} \]

### Final Answer
The volume of hydrogen gas evolved at STP is **1.12 liters**.

---