Question

N2 (g) + 3H2 (g)→2NH3 (g)
At a certain temperature and pressure, 1.6 L of N2 reacts with 4.8 L of H2.

Answer 1

The reaction you provided is the formation of ammonia (NH3) from nitrogen (N2) and hydrogen (H2):

\[ \text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g) \]

To determine how much of each reactant will be consumed in the reaction, we can use the stoichiometry of the balanced equation.

From the balanced equation, we can see that:

- 1 volume of N2 reacts with 3 volumes of H2.

Given:

- 1.6 L of N2
- 4.8 L of H2

### Determine the Required Amount of H2
Since 1 L of N2 requires 3 L of H2, we can calculate the amount of H2 required for 1.6 L of N2:

\[
\text{H2 required} = 1.6 \text{ L N2} \times 3 \text{ L H2}/1 \text{ L N2} = 4.8 \text{ L H2}
\]

### Analyze Reactants
- The amount of H2 available is exactly 4.8 L, which matches the amount needed for 1.6 L of N2.
- Therefore, both reactants will be completely consumed in this reaction.

### Product Formation
From the balanced equation, for every 1 L of N2 reacted, we yield 2 L of NH3:

\[
\text{NH3 produced} = 1.6 \text{ L N2} \times 2 \text{ L NH3}/1 \text{ L N2} = 3.2 \text{ L NH3}
\]

### Summary
- **Reactants consumed:**
- 1.6 L of N2
- 4.8 L of H2
- **Product formed:**
- 3.2 L of NH3

In conclusion, all of the N2 and H2 will be consumed, resulting in the production of 3.2 L of ammonia (NH3).