At S.T.P. a mixture of 280 mL of `CH_4 " and 140 mL of " H_2` is completely burnt. Calculate the required volume of oxygen and weight of water formed, assuming that whole of the steam condenses to water.

Explanation

To solve the problem of calculating the required volume of oxygen and the weight of water formed when a mixture of methane (CH₄) and hydrogen (H₂) is completely burnt at STP, we will follow these steps:

### Step 1: Write the Balanced Chemical Reactions
1. **Burning of Methane (CH₄)**:
\[ \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} \]
This indicates that 1 volume of methane reacts with 2 volumes of oxygen.

2. **Burning of Hydrogen (H₂)**:
\[ 2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O} \]
This indicates that 2 volumes of hydrogen react with 1 volume of oxygen.

### Step 2: Calculate the Volume of Oxygen Required for Each Reaction
1. **For Methane**:
- Given volume of CH₄ = 280 mL.
- According to the reaction, 1 volume of CH₄ requires 2 volumes of O₂.
- Therefore, the volume of O₂ required for CH₄:
\[ \text{Volume of O}_2 = 2 \times 280 \, \text{mL} = 560 \, \text{mL} \]

2. **For Hydrogen**:
- Given volume of H₂ = 140 mL.
- According to the reaction, 2 volumes of H₂ require 1 volume of O₂.
- Therefore, the volume of O₂ required for H₂:
\[ \text{Volume of O}_2 = \frac{1}{2} \times 140 \, \text{mL} = 70 \, \text{mL} \]

### Step 3: Calculate the Total Volume of Oxygen Required
- Total volume of O₂ required:
\[ \text{Total O}_2 = 560 \, \text{mL} + 70 \, \text{mL} = 630 \, \text{mL} \]

### Step 4: Calculate the Volume of Water Formed
1. **From Methane**:
- From the reaction, 1 volume of CH₄ produces 2 volumes of H₂O.
- Therefore, the volume of H₂O from CH₄:
\[ \text{Volume of H}_2\text{O} = 2 \times 280 \, \text{mL} = 560 \, \text{mL} \]

2. **From Hydrogen**:
- From the reaction, 2 volumes of H₂ produce 2 volumes of H₂O.
- Therefore, the volume of H₂O from H₂:
\[ \text{Volume of H}_2\text{O} = 140 \, \text{mL} \]

### Step 5: Calculate the Total Volume of Water Formed
- Total volume of H₂O formed:
\[ \text{Total H}_2\text{O} = 560 \, \text{mL} + 140 \, \text{mL} = 700 \, \text{mL} \]

### Step 6: Calculate the Weight of Water Formed
- Convert the volume of water to liters:
\[ \text{Volume of H}_2\text{O} = 700 \, \text{mL} = 0.700 \, \text{L} \]
- Using the molar volume of water (22.4 L/mol at STP):
\[ \text{Moles of H}_2\text{O} = \frac{0.700 \, \text{L}}{22.4 \, \text{L/mol}} = 0.03125 \, \text{mol} \]
- The molecular weight of water (H₂O) = 18 g/mol.
- Therefore, the weight of water formed:
\[ \text{Weight of H}_2\text{O} = 0.03125 \, \text{mol} \times 18 \, \text{g/mol} = 0.5625 \, \text{g} \]

### Final Answers
- Volume of oxygen required: **630 mL**
- Weight of water formed: **0.5625 g**