A mixture containing 100 gm H2 and 100 gm O2 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 -

Options

A. 112.5 gm
B. 50 gm
C. 25 gm
D. 200 gm

Answer

To solve the problem of how much water will be formed from the reaction of 100 grams of hydrogen (H₂) and 100 grams of oxygen (O₂), we can follow these steps:

### Step 1: Write the balanced chemical equation
The balanced equation for the reaction is:
\[ 2H_2 + O_2 \rightarrow 2H_2O \]

### Step 2: Calculate the molar masses
- The molar mass of \( H_2 \) (hydrogen) is \( 2 \, \text{g/mol} \) (since each H atom has a mass of 1 g/mol).
- The molar mass of \( O_2 \) (oxygen) is \( 32 \, \text{g/mol} \) (since each O atom has a mass of 16 g/mol).

### Step 3: Calculate the number of moles of each reactant
- For hydrogen:
\[ \text{Moles of } H_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{100 \, \text{g}}{2 \, \text{g/mol}} = 50 \, \text{moles} \]

- For oxygen:
\[ \text{Moles of } O_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{100 \, \text{g}}{32 \, \text{g/mol}} = 3.125 \, \text{moles} \]

### Step 4: Determine the limiting reactant
From the balanced equation, we see that 2 moles of \( H_2 \) react with 1 mole of \( O_2 \). Therefore, for 50 moles of \( H_2 \), we need:
\[ \text{Required moles of } O_2 = \frac{50 \, \text{moles of } H_2}{2} = 25 \, \text{moles of } O_2 \]
Since we only have 3.125 moles of \( O_2 \), oxygen is the limiting reactant.

### Step 5: Calculate the amount of water produced
According to the balanced equation, 1 mole of \( O_2 \) produces 2 moles of \( H_2O \). Therefore, 3.125 moles of \( O_2 \) will produce:
\[ \text{Moles of } H_2O = 3.125 \, \text{moles of } O_2 \times 2 = 6.25 \, \text{moles of } H_2O \]

### Step 6: Calculate the mass of water produced
The molar mass of water \( H_2O \) is \( 18 \, \text{g/mol} \). Thus, the mass of water produced is:
\[ \text{Mass of } H_2O = \text{moles} \times \text{molar mass} = 6.25 \, \text{moles} \times 18 \, \text{g/mol} = 112.5 \, \text{g} \]

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

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