For a reaction, `N_(2)(g)+3H_(2)(g)rarr2NH_(3)(g)`, identify dihydrogen `(H_(2))` as a limiting reagent in the following reaction mixtures.

Options

A. 28 g of `N_(2) + 6g` of `H_(2)`
B. 56 g of `N_(2) + 10 g` of `H_(2)`
C. 14 g of `N_(2) + 4g H_(2)`
D. 35g of `N_(2) + 8g` of `H_(2)`

Explanation

To identify dihydrogen (H₂) as a limiting reagent in the reaction \( N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \), we need to analyze the given quantities of nitrogen (N₂) and hydrogen (H₂) in different cases. The limiting reagent is the reactant that will be completely consumed first, thus limiting the amount of product formed.

### Step-by-Step Solution:

1. **Understand the Stoichiometry of the Reaction**:
- The balanced equation shows that 1 mole of N₂ reacts with 3 moles of H₂.
- Therefore, to completely react with 1 mole of N₂, we need 3 moles of H₂.

2. **Calculate the Molar Masses**:
- Molar mass of N₂ = 28 g/mol (since nitrogen has a molar mass of 14 g/mol).
- Molar mass of H₂ = 2 g/mol.

3. **Determine the Required Amount of H₂ for Given Amounts of N₂**:
- For 28 g of N₂, we require:
\[ \text{Required H₂} = \frac{3 \text{ moles H₂}}{1 \text{ mole N₂}} \times 28 \text{ g N₂} = 6 \text{ g H₂} \]
- This means that if we have 28 g of N₂, we need 6 g of H₂ to react completely.

4. **Analyze Each Case**:
- **Case 1**: 28 g of N₂ and 6 g of H₂
- Here, H₂ is not the limiting reagent since we have exactly the required amount.
- **Case 2**: 56 g of N₂ and 10 g of H₂
- For 56 g of N₂:
\[ \text{Required H₂} = \frac{6 \text{ g H₂}}{28 \text{ g N₂}} \times 56 \text{ g N₂} = 12 \text{ g H₂} \]
- Since we only have 10 g of H₂, H₂ is the limiting reagent.
- **Case 3**: 14 g of N₂ and 4 g of H₂
- For 14 g of N₂:
\[ \text{Required H₂} = \frac{6 \text{ g H₂}}{28 \text{ g N₂}} \times 14 \text{ g N₂} = 3 \text{ g H₂} \]
- Since we have 4 g of H₂, H₂ is not the limiting reagent.
- **Case 4**: 35 g of N₂ and 8 g of H₂
- For 35 g of N₂:
\[ \text{Required H₂} = \frac{6 \text{ g H₂}}{28 \text{ g N₂}} \times 35 \text{ g N₂} = 7.5 \text{ g H₂} \]
- Since we have 8 g of H₂, H₂ is not the limiting reagent.

5. **Conclusion**:
- The only case where H₂ is the limiting reagent is in Case 2 (56 g of N₂ and 10 g of H₂).