For the formation of 3.65g of HCI gas , what volume of hydrogen gas and chlorine gas , are required at NTP conditions?

Options

A. 1 L, 1L
B. 1.12 L, 2.24
C. 3.65 L, 1.83 L
D. 1.12 L, 1.12 L

Answer

To find the volume of hydrogen gas (H₂) and chlorine gas (Cl₂) required to produce 3.65 g of hydrochloric acid (HCl) at normal temperature and pressure (NTP), we can follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the formation of HCl from hydrogen and chlorine is:
\[ H_2 + Cl_2 \rightarrow 2 HCl \]

### Step 2: Calculate the molar mass of HCl
The molar mass of HCl can be calculated as follows:
- Molar mass of H = 1 g/mol
- Molar mass of Cl = 35.5 g/mol
- Therefore, molar mass of HCl = 1 + 35.5 = 36.5 g/mol

### Step 3: Calculate the number of moles of HCl produced
Using the mass of HCl given (3.65 g), we can calculate the number of moles of HCl:
\[ \text{Number of moles of HCl} = \frac{\text{mass}}{\text{molar mass}} = \frac{3.65 \, \text{g}}{36.5 \, \text{g/mol}} \approx 0.1 \, \text{mol} \]

### Step 4: Determine the stoichiometry of the reaction
From the balanced equation, we see that 1 mole of H₂ reacts with 1 mole of Cl₂ to produce 2 moles of HCl. Therefore, for every 2 moles of HCl produced, we need:
- 1 mole of H₂
- 1 mole of Cl₂

Thus, for 0.1 moles of HCl, we will need:
\[ \text{Moles of H}_2 = \frac{0.1 \, \text{mol HCl}}{2} = 0.05 \, \text{mol} \]
\[ \text{Moles of Cl}_2 = \frac{0.1 \, \text{mol HCl}}{2} = 0.05 \, \text{mol} \]

### Step 5: Calculate the volume of hydrogen and chlorine gases at NTP
At NTP, 1 mole of gas occupies 22.4 liters. Therefore, the volume of each gas can be calculated as follows:
\[ \text{Volume of H}_2 = 0.05 \, \text{mol} \times 22.4 \, \text{L/mol} = 1.12 \, \text{L} \]
\[ \text{Volume of Cl}_2 = 0.05 \, \text{mol} \times 22.4 \, \text{L/mol} = 1.12 \, \text{L} \]

### Final Answer
Thus, the volume of hydrogen gas and chlorine gas required to produce 3.65 g of HCl at NTP conditions is:
- Volume of H₂ = 1.12 L
- Volume of Cl₂ = 1.12 L