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

Answer

To determine when dihydrogen (H₂) acts as a limiting reagent in the reaction \(N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)\), we need to analyze the given reaction mixtures. 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 \(N_2\) reacts with 3 moles of \(H_2\). Therefore, for every mole of \(N_2\), we need 3 moles of \(H_2\).

2. **Calculate Moles of Reactants in Each Mixture**:
For each reaction mixture, we will convert the given mass of \(N_2\) and \(H_2\) into moles using the formula:
\[ \text{Number of moles} = \frac{\text{Given mass (g)}}{\text{Molar mass (g/mol)}} \]
- Molar mass of \(N_2\) = 28 g/mol
- Molar mass of \(H_2\) = 2 g/mol

3. **Analyze Each Mixture**:
- **Mixture 1**: 14 g of \(N_2\) and 4 g of \(H_2\)
- Moles of \(N_2\) = \( \frac{14}{28} = 0.5 \) moles
- Moles of \(H_2\) = \( \frac{4}{2} = 2 \) moles
- Required \(H_2\) for 0.5 moles of \(N_2\) = \(0.5 \times 3 = 1.5\) moles
- Since 2 moles of \(H_2\) are available, \(N_2\) is the limiting reagent.

- **Mixture 2**: 28 g of \(N_2\) and 6 g of \(H_2\)
- Moles of \(N_2\) = \( \frac{28}{28} = 1 \) mole
- Moles of \(H_2\) = \( \frac{6}{2} = 3 \) moles
- Required \(H_2\) for 1 mole of \(N_2\) = \(1 \times 3 = 3\) moles
- Both reactants are perfectly matched, so there is no limiting reagent.

- **Mixture 3**: 56 g of \(N_2\) and 10 g of \(H_2\)
- Moles of \(N_2\) = \( \frac{56}{28} = 2 \) moles
- Moles of \(H_2\) = \( \frac{10}{2} = 5 \) moles
- Required \(H_2\) for 2 moles of \(N_2\) = \(2 \times 3 = 6\) moles
- Since only 5 moles of \(H_2\) are available, \(H_2\) is the limiting reagent.

- **Mixture 4**: 35 g of \(N_2\) and 8 g of \(H_2\)
- Moles of \(N_2\) = \( \frac{35}{28} = 1.25 \) moles
- Moles of \(H_2\) = \( \frac{8}{2} = 4 \) moles
- Required \(H_2\) for 1.25 moles of \(N_2\) = \(1.25 \times 3 = 3.75\) moles
- Since 4 moles of \(H_2\) are available, \(N_2\) is the limiting reagent.

4. **Conclusion**:
In the third mixture, dihydrogen (H₂) acts as the limiting reagent.