10 moles of each `N_(2) and H_(2)` are made to react in a closed chamber. At equilibrium 40% of `H_(2)` was left. The total moles in the chamber are :

Options

A. 10
B. 12
C. 16
D. 8

Answer

To solve the problem step by step, we will analyze the reaction and the information given:

### Step 1: Write the balanced chemical equation
The balanced equation for the reaction between nitrogen and hydrogen to form ammonia is:
\[ N_2 + 3H_2 \rightarrow 2NH_3 \]

### Step 2: Determine initial moles
We are given that initially there are:
- 10 moles of \( N_2 \)
- 10 moles of \( H_2 \)

### Step 3: Calculate moles of hydrogen left at equilibrium
It is stated that at equilibrium, 40% of hydrogen is left. Therefore, we can calculate how much hydrogen is left:
\[ \text{Moles of } H_2 \text{ left} = 40\% \text{ of } 10 = \frac{40}{100} \times 10 = 4 \text{ moles} \]

### Step 4: Calculate moles of hydrogen reacted
To find out how many moles of hydrogen were consumed, we subtract the moles left from the initial moles:
\[ \text{Moles of } H_2 \text{ reacted} = 10 - 4 = 6 \text{ moles} \]

### Step 5: Determine moles of nitrogen consumed
From the balanced equation, we know that 3 moles of hydrogen react with 1 mole of nitrogen. Therefore, the moles of nitrogen consumed can be calculated as:
\[ \text{Moles of } N_2 \text{ consumed} = \frac{1}{3} \times \text{Moles of } H_2 \text{ reacted} = \frac{1}{3} \times 6 = 2 \text{ moles} \]

### Step 6: Calculate moles of ammonia produced
According to the balanced equation, for every 1 mole of nitrogen consumed, 2 moles of ammonia are produced. Thus, the moles of ammonia produced are:
\[ \text{Moles of } NH_3 \text{ produced} = 2 \times \text{Moles of } N_2 \text{ consumed} = 2 \times 2 = 4 \text{ moles} \]

### Step 7: Calculate total moles at equilibrium
Now we can find the total moles in the chamber at equilibrium:
- Moles of \( N_2 \) left = Initial moles - Moles consumed = \( 10 - 2 = 8 \) moles
- Moles of \( H_2 \) left = 4 moles (calculated previously)
- Moles of \( NH_3 \) produced = 4 moles

Thus, the total moles at equilibrium is:
\[ \text{Total moles} = \text{Moles of } N_2 + \text{Moles of } H_2 + \text{Moles of } NH_3 = 8 + 4 + 4 = 16 \text{ moles} \]

### Final Answer
The total moles in the chamber at equilibrium are **16 moles**.

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