3.0 g of `H_(2)` react with 29.0 g `O_(2)` to yield `H_(2)O`
(i) What is the limiting reactant ?
(ii) Calculate the maximum amount of water that can be formed
(iii) Calculate the amount of one of the reactants which remains unreacted.

Options

A. 6 g
B. 8 g
C. 5 .7 g
D. 5.2 g

Answer

To solve the problem step by step, we will follow the instructions provided in the question.

### Step 1: Write the balanced chemical equation
The reaction between hydrogen and oxygen to form water can be represented as:
\[ 2H_2 + O_2 \rightarrow 2H_2O \]

### Step 2: Calculate the moles of reactants
1. **Calculate moles of \( H_2 \)**:
- Molar mass of \( H_2 = 2 \, \text{g/mol} \)
- Moles of \( H_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{3.0 \, \text{g}}{2 \, \text{g/mol}} = 1.5 \, \text{mol} \)

2. **Calculate moles of \( O_2 \)**:
- Molar mass of \( O_2 = 32 \, \text{g/mol} \)
- Moles of \( O_2 = \frac{29.0 \, \text{g}}{32 \, \text{g/mol}} = 0.90625 \, \text{mol} \)

### Step 3: Determine the limiting reactant
From the balanced equation, we know that:
- 2 moles of \( H_2 \) react with 1 mole of \( O_2 \).

Using the moles calculated:
- Required moles of \( O_2 \) for 1.5 moles of \( H_2 \):
\[ \text{Required } O_2 = \frac{1.5 \, \text{mol} \, H_2}{2} = 0.75 \, \text{mol} \]

Since we have 0.90625 moles of \( O_2 \) available, which is more than the 0.75 moles required, \( H_2 \) is the limiting reactant.

### Step 4: Calculate the maximum amount of water that can be formed
From the balanced equation:
- 2 moles of \( H_2 \) produce 2 moles of \( H_2O \).

Thus, 1.5 moles of \( H_2 \) will produce:
\[ \text{Moles of } H_2O = 1.5 \, \text{mol} \]

Now, calculate the mass of water produced:
- Molar mass of \( H_2O = 18 \, \text{g/mol} \)
\[ \text{Mass of } H_2O = 1.5 \, \text{mol} \times 18 \, \text{g/mol} = 27 \, \text{g} \]

### Step 5: Calculate the amount of unreacted reactant
1. **Calculate the amount of \( O_2 \) consumed**:
- From the balanced equation, 1 mole of \( O_2 \) reacts with 2 moles of \( H_2 \).
- For 1.5 moles of \( H_2 \), the moles of \( O_2 \) consumed:
\[ \text{O}_2 \text{ consumed} = \frac{1.5 \, \text{mol} \, H_2}{2} = 0.75 \, \text{mol} \]

2. **Calculate the mass of \( O_2 \) consumed**:
\[ \text{Mass of } O_2 \text{ consumed} = 0.75 \, \text{mol} \times 32 \, \text{g/mol} = 24 \, \text{g} \]

3. **Calculate the unreacted \( O_2 \)**:
\[ \text{Unreacted } O_2 = 29.0 \, \text{g} - 24.0 \, \text{g} = 5.0 \, \text{g} \]

### Summary of Results
(i) The limiting reactant is \( H_2 \).
(ii) The maximum amount of water that can be formed is \( 27 \, \text{g} \).
(iii) The amount of \( O_2 \) that remains unreacted is \( 5.0 \, \text{g} \).