How many litres of NH3 would be formed when 120 L of `H_(2)`, at standard conditions combine with `N_(2)`? What is the volume of `N_(2)` consumed in this reaction?

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical equation
The formation of ammonia (NH3) from nitrogen (N2) and hydrogen (H2) can be represented by the following balanced equation:

\[ N_2(g) + 3H_2(g) \leftrightarrow 2NH_3(g) \]

### Step 2: Identify the volumes involved
We are given that 120 L of hydrogen (H2) is used in the reaction. According to the balanced equation, 3 volumes of hydrogen react with 1 volume of nitrogen to produce 2 volumes of ammonia.

### Step 3: Calculate the volume of ammonia produced
Using the stoichiometric ratios from the balanced equation, we can find out how much ammonia (NH3) is produced from 120 L of hydrogen (H2).

From the equation:
- 3 volumes of H2 produce 2 volumes of NH3.

Using the ratio:
\[ \text{Volume of } NH_3 = \left(\frac{2 \text{ L of } NH_3}{3 \text{ L of } H_2}\right) \times 120 \text{ L of } H_2 \]
\[ \text{Volume of } NH_3 = \frac{2}{3} \times 120 = 80 \text{ L} \]

### Step 4: Calculate the volume of nitrogen consumed
Next, we need to find out how much nitrogen (N2) is consumed in the reaction. According to the balanced equation:
- 1 volume of N2 reacts with 3 volumes of H2.

Using the ratio:
\[ \text{Volume of } N_2 = \left(\frac{1 \text{ L of } N_2}{3 \text{ L of } H_2}\right) \times 120 \text{ L of } H_2 \]
\[ \text{Volume of } N_2 = \frac{120}{3} = 40 \text{ L} \]

### Final Answers:
- The volume of ammonia (NH3) produced is **80 L**.
- The volume of nitrogen (N2) consumed is **40 L**.

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