What is the equivalent weight of `NH_(3)` in the given reaction?
`3CuO+2NH_(3) rarr 3Cu+N_(2)+3H_(2)O`

Options

A. `17`
B. `(17)/(4)`
C. `(17)/(2)`
D. `(17)/(3)`

Explanation

To find the equivalent weight of \( NH_3 \) in the given reaction:

\[ 3CuO + 2NH_3 \rightarrow 3Cu + N_2 + 3H_2O \]

we will follow these steps:

### Step 1: Identify the change in oxidation states
In the reaction, we need to determine how many electrons are transferred when \( NH_3 \) is oxidized to \( N_2 \).

- In \( NH_3 \), nitrogen has an oxidation state of -3.
- In \( N_2 \), nitrogen has an oxidation state of 0.

### Step 2: Calculate the change in oxidation state
The change in oxidation state for one nitrogen atom is:
\[ 0 - (-3) = 3 \]
Since there are 2 nitrogen atoms in \( 2NH_3 \), the total change in oxidation state is:
\[ 2 \times 3 = 6 \]

### Step 3: Determine the number of moles of \( NH_3 \)
The balanced equation shows that 2 moles of \( NH_3 \) are used in the reaction.

### Step 4: Calculate the balance factor
The balance factor is defined as the total number of electrons transferred divided by the number of moles of the substance involved. Therefore, for \( NH_3 \):
\[ \text{Balance Factor} = \frac{\text{Total electrons transferred}}{\text{Number of moles of } NH_3} = \frac{6}{2} = 3 \]

### Step 5: Calculate the molecular weight of \( NH_3 \)
The molecular weight of \( NH_3 \) is calculated as follows:
- Nitrogen (N) = 14 g/mol
- Hydrogen (H) = 1 g/mol (and there are 3 hydrogen atoms)

Thus, the molecular weight of \( NH_3 \) is:
\[ 14 + (3 \times 1) = 17 \text{ g/mol} \]

### Step 6: Calculate the equivalent weight
The equivalent weight is given by the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molecular Weight}}{\text{Balance Factor}} = \frac{17}{3} \approx 5.67 \text{ g/equiv} \]

### Final Answer
The equivalent weight of \( NH_3 \) in the given reaction is approximately \( 5.67 \text{ g/equiv} \).

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