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The problem asks to determine the volume of water produced in a reaction where ammonia gas reacts with oxygen gas, given that 8.6 m³ of oxygen were consumed.

The balanced chemical equation for the reaction of ammonia (NH₃) with oxygen (O₂) is:

\[
4 \text{NH}_3 + 5 \text{O}_2 \rightarrow 4 \text{NO} + 6 \text{H}_2\text{O}
\]

From the balanced equation, for every 5 volumes of oxygen (O₂) consumed, 6 volumes of water vapor (H₂O) are produced. The ratio of oxygen to water vapor is:

\[
\frac{6 \text{H}_2\text{O}}{5 \text{O}_2}
\]

Using this ratio, we can calculate the volume of water produced when 8.6 m³ of oxygen is consumed:

\[
\text{Volume of water} = 8.6 \, \text{m}^3 \times \frac{6 \text{H}_2\text{O}}{5 \text{O}_2}
\]

Now let's calculate this.The volume of water vapor produced is approximately **10.32 m³** when 8.6 m³ of oxygen is consumed, rounded to the correct number of significant digits.

Let me know if you'd like more details, and here are some related questions:

1. What is the stoichiometric ratio between ammonia and oxygen in this reaction?
2. How would the volume of water change if 10 m³ of oxygen were used?
3. Why is it important to balance chemical equations before performing volume calculations?
4. What conditions (e.g., temperature, pressure) would affect the gas volumes in this reaction?
5. How can you determine the mass of water produced from the volume of water vapor?

**Tip**: When working with gas volumes, ensure you are consistent with temperature and pressure conditions, as gases behave differently under varying conditions (use of the ideal gas law might help in more complex situations).

Ammonia gas and oxygen gas react to form water vapor and nitrogen monoxide gas. What volume of water would be produced by this reaction if 8.6 m³ of oxygen were consumed?

Calculate the volume of water vapor produced in a reaction between ammonia and oxygen gas, given 8.6 m³ of oxygen is consumed. Learn about stoichiometry and gas volume relationships in this high school chemistry problem.

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