In which of the following minimum amount of `O_(2)` is required per gram of reactant

Options

A. `4Fe(s) + 3O_(2)(g) rarr 2Fe_(2)O_(3)(s)`
B. `2M g(s) + O_(2)(g) rarr 2Mg O(s)`
C. `P_(4)(s) + 5O_(2)(g) rarr P_(4)O_(10)(s)`
D. `C_(3)H_(8)(g) + 5O_(2)(g) rarr 3CO_(2)(g) + 4H_(2)O (l)`

Explanation

To determine which reactant requires the minimum amount of \( O_2 \) per gram, we will analyze each option step by step.

### Step 1: Analyze Option A - Iron Reaction
1. **Reaction**: \( 4 \text{Fe} + 3 \text{O}_2 \rightarrow 2 \text{Fe}_2\text{O}_3 \)
2. **Molecular Weight of Iron (Fe)**: 56 g/mol
3. **Moles of Reactant (Fe)**: 4 moles
4. **Molecular Weight of Oxygen (O2)**: 32 g/mol
5. **Moles of Oxygen**: 3 moles

Using the formula for the minimum amount of \( O_2 \) required per gram of reactant:
\[ \text{Amount of } O_2 = \frac{\text{Number of moles of } O_2 \times \text{Molecular weight of } O_2}{\text{Number of moles of Fe} \times \text{Molecular weight of Fe}} \]
Substituting the values:
\[ \text{Amount of } O_2 = \frac{3 \times 32}{4 \times 56} = \frac{96}{224} = 0.4286 \text{ g} \]

### Step 2: Analyze Option B - Magnesium Reaction
1. **Reaction**: \( 2 \text{Mg} + \text{O}_2 \rightarrow 2 \text{MgO} \)
2. **Molecular Weight of Magnesium (Mg)**: 24 g/mol
3. **Moles of Reactant (Mg)**: 2 moles
4. **Molecular Weight of Oxygen (O2)**: 32 g/mol
5. **Moles of Oxygen**: 1 mole

Using the same formula:
\[ \text{Amount of } O_2 = \frac{1 \times 32}{2 \times 24} = \frac{32}{48} = 0.6667 \text{ g} \]

### Step 3: Analyze Option C - Phosphorus Reaction
1. **Reaction**: \( \text{P}_4 + 5 \text{O}_2 \rightarrow \text{P}_4\text{O}_{10} \)
2. **Molecular Weight of Phosphorus (P4)**: 124 g/mol
3. **Moles of Reactant (P4)**: 1 mole
4. **Molecular Weight of Oxygen (O2)**: 32 g/mol
5. **Moles of Oxygen**: 5 moles

Using the formula:
\[ \text{Amount of } O_2 = \frac{5 \times 32}{1 \times 124} = \frac{160}{124} = 1.2903 \text{ g} \]

### Step 4: Analyze Option D - Propane Reaction
1. **Reaction**: \( \text{C}_3\text{H}_8 + 5 \text{O}_2 \rightarrow 3 \text{CO}_2 + 4 \text{H}_2\text{O} \)
2. **Molecular Weight of Propane (C3H8)**: 44 g/mol
3. **Moles of Reactant (C3H8)**: 1 mole
4. **Molecular Weight of Oxygen (O2)**: 32 g/mol
5. **Moles of Oxygen**: 5 moles

Using the formula:
\[ \text{Amount of } O_2 = \frac{5 \times 32}{1 \times 44} = \frac{160}{44} = 3.6364 \text{ g} \]

### Conclusion
Now, we compare the amounts of \( O_2 \) required for each reactant:
- Iron: 0.4286 g
- Magnesium: 0.6667 g
- Phosphorus: 1.2903 g
- Propane: 3.6364 g

The minimum amount of \( O_2 \) required per gram of reactant is for Iron, which is **0.4286 g**.

### Final Answer
**Option A (Iron) requires the minimum amount of \( O_2 \).**