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 which reagent is the limiting reagent in the reaction \( N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \), we need to analyze the given quantities of \( N_2 \) and \( H_2 \) in the reaction mixtures.

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equation**:
The balanced equation for the reaction is:
\[ N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \]

2. **Calculate Moles of Each Reactant**:
- For \( N_2 \):
- Given mass = 56 g
- Molar mass of \( N_2 \) = 28 g/mol
- Moles of \( N_2 \) = \( \frac{56 \text{ g}}{28 \text{ g/mol}} = 2 \text{ moles} \)

- For \( H_2 \):
- Given mass = 10 g
- Molar mass of \( H_2 \) = 2 g/mol
- Moles of \( H_2 \) = \( \frac{10 \text{ g}}{2 \text{ g/mol}} = 5 \text{ moles} \)

3. **Determine Stoichiometric Ratios**:
From the balanced equation, the stoichiometric ratio of \( N_2 \) to \( H_2 \) is 1:3. This means:
- 1 mole of \( N_2 \) reacts with 3 moles of \( H_2 \).

4. **Calculate the Required Moles of \( H_2 \)**:
To find out how much \( H_2 \) is required for the available \( N_2 \):
- For 2 moles of \( N_2 \), the required moles of \( H_2 \) = \( 2 \text{ moles } N_2 \times 3 \text{ moles } H_2/\text{mole } N_2 = 6 \text{ moles } H_2 \).

5. **Compare Available Moles of \( H_2 \) with Required Moles**:
- Available moles of \( H_2 \) = 5 moles
- Required moles of \( H_2 \) = 6 moles

6. **Identify the Limiting Reagent**:
Since we have only 5 moles of \( H_2 \) available, which is less than the 6 moles required, \( H_2 \) is the limiting reagent.

### Conclusion:
In the given reaction mixture, \( H_2 \) is identified as the limiting reagent.