A mixture containing 100 gm `H_(2)` and 100 gm `O_(2)` is ignited so that water is formed according to the reaction, `2H_(2)+O_(2) rarr 2H_(2)O`, How much water will be formed -

Answer

To solve the problem step by step, we will follow the stoichiometric calculations based on the balanced chemical reaction provided.

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equation**:
The reaction for the formation of water from hydrogen and oxygen is:
\[ 2H_2 + O_2 \rightarrow 2H_2O \]

2. **Calculate the Moles of Reactants**:
- **For Hydrogen (H₂)**:
- Given mass = 100 g
- Molar mass of H₂ = 2 g/mol (1 g/mol for H × 2)
- Moles of H₂ = \(\frac{100 \text{ g}}{2 \text{ g/mol}} = 50 \text{ moles}\)

- **For Oxygen (O₂)**:
- Given mass = 100 g
- Molar mass of O₂ = 32 g/mol (16 g/mol for O × 2)
- Moles of O₂ = \(\frac{100 \text{ g}}{32 \text{ g/mol}} = 3.125 \text{ moles}\)

3. **Determine the Limiting Reactant**:
- According to the balanced equation, 2 moles of H₂ react with 1 mole of O₂.
- Therefore, for 50 moles of H₂, the required moles of O₂ = \(\frac{50 \text{ moles H₂}}{2} = 25 \text{ moles O₂}\).
- We only have 3.125 moles of O₂ available, which is less than the required 25 moles.
- Thus, O₂ is the limiting reactant.

4. **Calculate the Moles of Water (H₂O) Produced**:
- From the balanced equation, 1 mole of O₂ produces 2 moles of H₂O.
- Therefore, 3.125 moles of O₂ will produce:
\[ 3.125 \text{ moles O₂} \times 2 = 6.25 \text{ moles H₂O} \]

5. **Calculate the Mass of Water Produced**:
- Molar mass of H₂O = 18 g/mol (2 g/mol for H × 2 + 16 g/mol for O)
- Mass of H₂O = Moles of H₂O × Molar mass of H₂O
\[ \text{Mass of H₂O} = 6.25 \text{ moles} \times 18 \text{ g/mol} = 112.5 \text{ g} \]

### Final Answer:
The mass of water formed is **112.5 grams**.

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