`{:(,"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 reactions in Column I with the corresponding mass of products in Column II, we will analyze each reaction step by step.

### Step-by-Step Solution:

**Step 1: Analyze Reaction A**
- **Reaction:** \(2H_2 + O_2 \rightarrow 2H_2O\)
- **Given:** 1 g of \(H_2\) and 1 g of \(O_2\)
- **Molar Masses:**
- \(H_2 = 2 \, \text{g/mol}\)
- \(O_2 = 32 \, \text{g/mol}\)
- **Moles Calculation:**
- Moles of \(H_2 = \frac{1 \, \text{g}}{2 \, \text{g/mol}} = 0.5 \, \text{mol}\)
- Moles of \(O_2 = \frac{1 \, \text{g}}{32 \, \text{g/mol}} \approx 0.03125 \, \text{mol}\)
- **Limiting Reagent:** \(O_2\) (since it is less in moles)
- **Mass of \(H_2O\) produced:**
- Moles of \(H_2O = 2 \times \text{moles of } O_2 = 2 \times 0.03125 = 0.0625 \, \text{mol}\)
- Mass of \(H_2O = 0.0625 \, \text{mol} \times 18 \, \text{g/mol} = 1.125 \, \text{g}\)

**Step 2: Analyze Reaction B**
- **Reaction:** \(3H_2 + N_2 \rightarrow 2NH_3\)
- **Given:** 1 g of \(H_2\) and 1 g of \(N_2\)
- **Molar Masses:**
- \(N_2 = 28 \, \text{g/mol}\)
- **Moles Calculation:**
- Moles of \(H_2 = 0.5 \, \text{mol}\) (as calculated before)
- Moles of \(N_2 = \frac{1 \, \text{g}}{28 \, \text{g/mol}} \approx 0.0357 \, \text{mol}\)
- **Limiting Reagent:** \(N_2\)
- **Mass of \(NH_3\) produced:**
- Moles of \(NH_3 = 2 \times \text{moles of } N_2 = 2 \times 0.0357 \approx 0.0714 \, \text{mol}\)
- Mass of \(NH_3 = 0.0714 \, \text{mol} \times 17 \, \text{g/mol} \approx 1.214 \, \text{g}\)

**Step 3: Analyze Reaction C**
- **Reaction:** \(H_2 + Cl_2 \rightarrow 2HCl\)
- **Given:** 1 g of \(H_2\) and 1 g of \(Cl_2\)
- **Molar Masses:**
- \(Cl_2 = 71 \, \text{g/mol}\)
- **Moles Calculation:**
- Moles of \(H_2 = 0.5 \, \text{mol}\)
- Moles of \(Cl_2 = \frac{1 \, \text{g}}{71 \, \text{g/mol}} \approx 0.0141 \, \text{mol}\)
- **Limiting Reagent:** \(Cl_2\)
- **Mass of \(HCl\) produced:**
- Moles of \(HCl = 2 \times \text{moles of } Cl_2 = 2 \times 0.0141 \approx 0.0282 \, \text{mol}\)
- Mass of \(HCl = 0.0282 \, \text{mol} \times 36.5 \, \text{g/mol} \approx 1.028 \, \text{g}\)

**Step 4: Analyze Reaction D**
- **Reaction:** \(2H_2 + C \rightarrow CH_4\)
- **Given:** 1 g of \(H_2\) and 1 g of \(C\)
- **Molar Masses:**
- \(C = 12 \, \text{g/mol}\)
- **Moles Calculation:**
- Moles of \(H_2 = 0.5 \, \text{mol}\)
- Moles of \(C = \frac{1 \, \text{g}}{12 \, \text{g/mol}} \approx 0.0833 \, \text{mol}\)
- **Limiting Reagent:** \(H_2\)
- **Mass of \(CH_4\) produced:**
- Moles of \(CH_4 = \text{moles of } H_2 = 0.5 \, \text{mol}\)
- Mass of \(CH_4 = 0.5 \, \text{mol} \times 16 \, \text{g/mol} = 1.33 \, \text{g}\)

### Final Matching:
- 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 (1.33 g)