3.0 g of `H_(2)` react with 29.0 g of `O_(2)` yield `H_(2)O`.
(i) Which is the limiting reagent.
(ii) Calculate the maximum amount of `H_(2)O` that can be formed
(iii) Calculate the amount of reactant left unreacted

Answer

To solve the problem step by step, we will follow the given parts of the question:

### Step 1: Write the balanced chemical equation
The reaction between hydrogen and oxygen to form water is given by the equation:
\[ 2H_2 + O_2 \rightarrow 2H_2O \]

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

2. **Moles of \( O_2 \)**:
- Molar mass of \( O_2 \) = 32 g/mol
- Given mass of \( O_2 \) = 29.0 g
\[ \text{Moles of } O_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{29.0 \, \text{g}}{32 \, \text{g/mol}} = 0.90625 \, \text{moles} \approx 0.9 \, \text{moles} \]

### Step 3: Determine the limiting reagent
From the balanced equation, we see that:
- 2 moles of \( H_2 \) react with 1 mole of \( O_2 \).
- Therefore, 1.5 moles of \( H_2 \) would require:
\[ \text{Required moles of } O_2 = \frac{1.5 \, \text{moles of } H_2}{2} = 0.75 \, \text{moles of } O_2 \]

Since we have 0.9 moles of \( O_2 \) available, which is more than the 0.75 moles required, \( H_2 \) is the limiting reagent.

### Step 4: Calculate the maximum amount of \( H_2O \) formed
According to the balanced equation:
- 2 moles of \( H_2 \) produce 2 moles of \( H_2O \).
- Therefore, 1.5 moles of \( H_2 \) will produce:
\[ \text{Moles of } H_2O = 1.5 \, \text{moles of } H_2 = 1.5 \, \text{moles of } H_2O \]

Now, to find the mass of \( H_2O \):
- Molar mass of \( H_2O \) = 18 g/mol
\[ \text{Mass of } H_2O = 1.5 \, \text{moles} \times 18 \, \text{g/mol} = 27.0 \, \text{g} \]

### Step 5: Calculate the amount of reactant left unreacted
We have 0.9 moles of \( O_2 \) initially, and we used 0.75 moles:
\[ \text{Unreacted } O_2 = 0.9 \, \text{moles} - 0.75 \, \text{moles} = 0.15 \, \text{moles} \]

Now, to find the mass of unreacted \( O_2 \):
\[ \text{Mass of unreacted } O_2 = 0.15 \, \text{moles} \times 32 \, \text{g/mol} = 4.8 \, \text{g} \]

### Final Answers:
(i) Limiting reagent: \( H_2 \)
(ii) Maximum amount of \( H_2O \) formed: 27.0 g
(iii) Amount of \( O_2 \) left unreacted: 4.8 g