For the following reaction
`N_(2)+3H_(2)to2NH_(3)` equivalent mass of `N_(2)=("molar mass of" N_(2))/X`
What is the value of x.

Answer

To find the value of \( x \) in the expression for the equivalent mass of \( N_2 \), we can follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction is:
\[ N_2 + 3H_2 \rightarrow 2NH_3 \]

### Step 2: Identify the oxidation states
In the reaction:
- The oxidation state of nitrogen in \( N_2 \) is 0.
- The oxidation state of nitrogen in \( NH_3 \) is -3.
- The oxidation state of hydrogen in \( H_2 \) is 0.
- The oxidation state of hydrogen in \( NH_3 \) is +1.

### Step 3: Calculate the change in oxidation state
For each nitrogen atom, the change in oxidation state from 0 in \( N_2 \) to -3 in \( NH_3 \) is:
\[ \Delta \text{oxidation state} = 0 - (-3) = 3 \]
Since there are 2 nitrogen atoms in the product \( 2NH_3 \), the total change for nitrogen is:
\[ \text{Total change} = 2 \times 3 = 6 \]

### Step 4: Determine the number of equivalents
The number of equivalents for \( N_2 \) can be calculated based on the total change in oxidation state. Since the total change is 6, we can say that:
\[ \text{Number of equivalents} = 6 \]

### Step 5: Calculate the molar mass of \( N_2 \)
The molar mass of nitrogen (\( N_2 \)) is:
\[ \text{Molar mass of } N_2 = 14 \, \text{g/mol} \times 2 = 28 \, \text{g/mol} \]

### Step 6: Calculate the equivalent mass
The equivalent mass of \( N_2 \) is given by the formula:
\[ \text{Equivalent mass} = \frac{\text{Molar mass}}{X} \]
From the previous steps, we know that the number of equivalents (X) is 6. Therefore, we can write:
\[ \text{Equivalent mass of } N_2 = \frac{28 \, \text{g/mol}}{6} \]

### Step 7: Solve for \( x \)
From the equivalent mass formula, we have:
\[ X = 6 \]

Thus, the value of \( x \) is:
\[ \boxed{6} \]