`{:(,"Column-I",,"Column-II",),(,,,("mass of product"),),((A),2H_(2)+O_(2)rarr2H_(2)O,(p),1.028 g,),(,lg" "lg,,,),((B),3H_(2) +N_(2) rarr2NH_(3),(q),1.333 g,),(,lg " " lg,,,),((C),H_(2) + CI_(2) rarr 2HCI ,(r),1.125 g,),(,lg " "lg ,,,),((D),2H_(2) + C rarrCH_(4),(s),1.214 g,),(,lg " "lg,,,):}`

Answer

To solve the problem of matching the chemical equations with their respective mass of products, we will analyze each reaction step by step.

### Step 1: Analyze Reaction A
**Reaction:** \( 2H_2 + O_2 \rightarrow 2H_2O \)

1. **Molar Mass Calculation:**
- Molar mass of \( H_2 = 2 \, \text{g/mol} \)
- Molar mass of \( O_2 = 32 \, \text{g/mol} \)
- Molar mass of \( H_2O = 18 \, \text{g/mol} \)

2. **Assuming 1 g of \( O_2 \):**
- Moles of \( O_2 = \frac{1 \, \text{g}}{32 \, \text{g/mol}} = \frac{1}{32} \, \text{mol} \)

3. **Calculate Moles of \( H_2O \):**
- From the reaction, \( 1 \, \text{mol} \, O_2 \) produces \( 2 \, \text{mol} \, H_2O \).
- Therefore, \( \frac{1}{32} \, \text{mol} \, O_2 \) produces \( 2 \times \frac{1}{32} = \frac{1}{16} \, \text{mol} \, H_2O \).

4. **Calculate Mass of \( H_2O \):**
- Mass of \( H_2O = \text{moles} \times \text{molar mass} = \frac{1}{16} \times 18 = 1.125 \, \text{g} \)

**Match:** A matches with (r) 1.125 g.

### Step 2: Analyze Reaction B
**Reaction:** \( 3H_2 + N_2 \rightarrow 2NH_3 \)

1. **Assuming 1 g of \( N_2 \):**
- Moles of \( N_2 = \frac{1 \, \text{g}}{28 \, \text{g/mol}} = \frac{1}{28} \, \text{mol} \)

2. **Calculate Moles of \( NH_3 \):**
- From the reaction, \( 1 \, \text{mol} \, N_2 \) produces \( 2 \, \text{mol} \, NH_3 \).
- Therefore, \( \frac{1}{28} \, \text{mol} \, N_2 \) produces \( 2 \times \frac{1}{28} = \frac{1}{14} \, \text{mol} \, NH_3 \).

3. **Calculate Mass of \( NH_3 \):**
- Molar mass of \( NH_3 = 17 \, \text{g/mol} \)
- Mass of \( NH_3 = \frac{1}{14} \times 17 = 1.214 \, \text{g} \)

**Match:** B matches with (s) 1.214 g.

### Step 3: Analyze Reaction C
**Reaction:** \( H_2 + Cl_2 \rightarrow 2HCl \)

1. **Assuming 1 g of \( Cl_2 \):**
- Moles of \( Cl_2 = \frac{1 \, \text{g}}{71 \, \text{g/mol}} = \frac{1}{71} \, \text{mol} \)

2. **Calculate Moles of \( HCl \):**
- From the reaction, \( 1 \, \text{mol} \, Cl_2 \) produces \( 2 \, \text{mol} \, HCl \).
- Therefore, \( \frac{1}{71} \, \text{mol} \, Cl_2 \) produces \( 2 \times \frac{1}{71} = \frac{2}{71} \, \text{mol} \, HCl \).

3. **Calculate Mass of \( HCl \):**
- Molar mass of \( HCl = 36.5 \, \text{g/mol} \)
- Mass of \( HCl = \frac{2}{71} \times 36.5 = 1.028 \, \text{g} \)

**Match:** C matches with (p) 1.028 g.

### Step 4: Analyze Reaction D
**Reaction:** \( 2H_2 + C \rightarrow CH_4 \)

1. **Assuming 1 g of \( C \):**
- Moles of \( C = \frac{1 \, \text{g}}{12 \, \text{g/mol}} = \frac{1}{12} \, \text{mol} \)

2. **Calculate Moles of \( CH_4 \):**
- From the reaction, \( 1 \, \text{mol} \, C \) produces \( 1 \, \text{mol} \, CH_4 \).
- Therefore, \( \frac{1}{12} \, \text{mol} \, C \) produces \( \frac{1}{12} \, \text{mol} \, CH_4 \).

3. **Calculate Mass of \( CH_4 \):**
- Molar mass of \( CH_4 = 16 \, \text{g/mol} \)
- Mass of \( CH_4 = \frac{1}{12} \times 16 = 1.333 \, \text{g} \)

**Match:** D matches with (q) 1.333 g.

### Summary of Matches:
- A matches with (r) 1.125 g
- B matches with (s) 1.214 g
- C matches with (p) 1.028 g
- D matches with (q) 1.333 g