6 g of `H_(2)` reacts with 14 g `N_(2)` to form `NH_(3)` till the reaction completely consumes the limiting reagent. The mass of other reactant (in g) left are ……

Explanation

To solve the problem step by step, we will follow these procedures:

### Step 1: Write the balanced chemical equation.
The reaction between hydrogen (H₂) and nitrogen (N₂) to form ammonia (NH₃) can be represented as:
\[ N_2 + 3H_2 \rightarrow 2NH_3 \]

### Step 2: Calculate the number of moles of each reactant.
1. For hydrogen (H₂):
- Given mass of H₂ = 6 g
- Molecular weight of H₂ = 2 g/mol
\[ \text{Number of moles of } H_2 = \frac{\text{mass}}{\text{molecular weight}} = \frac{6 \, \text{g}}{2 \, \text{g/mol}} = 3 \, \text{moles} \]

2. For nitrogen (N₂):
- Given mass of N₂ = 14 g
- Molecular weight of N₂ = 28 g/mol
\[ \text{Number of moles of } N_2 = \frac{\text{mass}}{\text{molecular weight}} = \frac{14 \, \text{g}}{28 \, \text{g/mol}} = 0.5 \, \text{moles} \]

### Step 3: Determine the limiting reagent.
To find the limiting reagent, we compare the mole ratio of the reactants based on the balanced equation:
- From the balanced equation, 1 mole of N₂ reacts with 3 moles of H₂.
- For 0.5 moles of N₂, the required moles of H₂ would be:
\[ 0.5 \, \text{moles of } N_2 \times 3 \, \text{moles of } H_2 = 1.5 \, \text{moles of } H_2 \]
- We have 3 moles of H₂ available, which is more than enough to react with 0.5 moles of N₂.

Thus, **N₂ is the limiting reagent**.

### Step 4: Calculate the amount of H₂ that reacts.
Since N₂ is the limiting reagent, we can calculate how much H₂ will react with it:
- 0.5 moles of N₂ will react with 1.5 moles of H₂.

### Step 5: Calculate the remaining amount of H₂.
- Initial moles of H₂ = 3 moles
- Moles of H₂ that reacted = 1.5 moles
- Remaining moles of H₂:
\[ \text{Remaining } H_2 = 3 \, \text{moles} - 1.5 \, \text{moles} = 1.5 \, \text{moles} \]

### Step 6: Convert the remaining moles of H₂ to grams.
- To find the mass of the remaining H₂:
\[ \text{Mass of remaining } H_2 = \text{remaining moles} \times \text{molecular weight} = 1.5 \, \text{moles} \times 2 \, \text{g/mol} = 3 \, \text{g} \]

### Final Answer:
The mass of the other reactant (H₂) left is **3 g**.

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