The mass of `Mg_(3)N_(2)` produced if 48 gm metal is reacted with 34 gm `NH_(3)` gas is
Mg + `NH_(3)` `to` `Mg_(3)N_(2)` + `H_(2)`

Options

A. 200/3 gm
B. 100/3 gm
C. 400/3 gm
D. 150/3 gm

Answer

To find the mass of \( \text{Mg}_3\text{N}_2 \) produced when 48 g of magnesium reacts with 34 g of ammonia, we can follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction is:
\[ 3 \text{Mg} + 2 \text{NH}_3 \rightarrow \text{Mg}_3\text{N}_2 + 3 \text{H}_2 \]

### Step 2: Calculate the moles of each reactant
- **Moles of Magnesium (Mg)**:
The molar mass of magnesium (Mg) is 24 g/mol.
\[ \text{Moles of Mg} = \frac{\text{mass}}{\text{molar mass}} = \frac{48 \text{ g}}{24 \text{ g/mol}} = 2 \text{ moles} \]

- **Moles of Ammonia (NH3)**:
The molar mass of ammonia (NH3) is 17 g/mol.
\[ \text{Moles of NH}_3 = \frac{34 \text{ g}}{17 \text{ g/mol}} = 2 \text{ moles} \]

### Step 3: Determine the limiting reagent
From the balanced equation, we see that:
- 3 moles of Mg react with 2 moles of NH3.

Now, we can compare the mole ratio:
- For 2 moles of NH3, we would need \( \frac{3}{2} \times 2 = 3 \) moles of Mg.
- We only have 2 moles of Mg available.

Thus, magnesium (Mg) is the limiting reagent.

### Step 4: Calculate the moles of \( \text{Mg}_3\text{N}_2 \) produced
According to the balanced equation:
- 3 moles of Mg produce 1 mole of \( \text{Mg}_3\text{N}_2 \).

So, from 2 moles of Mg:
\[ \text{Moles of } \text{Mg}_3\text{N}_2 = \frac{2 \text{ moles Mg}}{3} = \frac{2}{3} \text{ moles} \]

### Step 5: Calculate the molar mass of \( \text{Mg}_3\text{N}_2 \)
The molar mass of \( \text{Mg}_3\text{N}_2 \) is calculated as follows:
- Molar mass of Mg = 24 g/mol
- Molar mass of N = 14 g/mol

Thus,
\[ \text{Molar mass of } \text{Mg}_3\text{N}_2 = (3 \times 24) + (2 \times 14) = 72 + 28 = 100 \text{ g/mol} \]

### Step 6: Calculate the mass of \( \text{Mg}_3\text{N}_2 \) produced
Using the number of moles calculated:
\[ \text{Mass of } \text{Mg}_3\text{N}_2 = \text{moles} \times \text{molar mass} = \frac{2}{3} \text{ moles} \times 100 \text{ g/mol} = \frac{200}{3} \text{ g} \approx 66.67 \text{ g} \]

### Final Answer
The mass of \( \text{Mg}_3\text{N}_2 \) produced is approximately 66.67 g.

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