What volume of carbon (ll) oxide is produced by reacting excess carbon with 10 dm3 of oxygen?

1. Analyze the question

We are asked to find the volume of carbon(II) oxide ($CO$) produced when a specific volume of oxygen ($O_2$) reacts with excess carbon ($C$). This is a stoichiometry problem based on the Law of Combining Volumes (Gay-Lussac's Law).

2. Explain the concept

According to Gay-Lussac's Law of Combining Volumes, when gases react, they do so in volumes which are simple ratios to one another and to the volume of the products, provided the temperature and pressure remain constant.

First, we must write the balanced chemical equation for the reaction:

From the balanced equation, the mole ratio of $O_2$ to $CO$ is 1 : 2. This means that 1 volume of oxygen gas will produce 2 volumes of carbon(II) oxide gas.

3. Show the solution

• Step 1: Identify the mole ratio from the balanced equation. $$1 \text{ mole of } O_2 \text{ produces } 2 \text{ moles of } CO$$
• Step 2: Apply the ratio to the given volume. Since the ratio is 1:2, then: $$1 \text{ dm}^3 \text{ of } O_2 \text{ produces } 2 \text{ dm}^3 \text{ of } CO$$
• Step 3: Calculate for 10 dm³ of oxygen. $$10 \text{ dm}^3 \text{ of } O_2 \times \left( \frac{2 \text{ dm}^3 \text{ of } CO}{1 \text{ dm}^3 \text{ of } O_2} \right) = 20 \text{ dm}^3 \text{ of } CO$$

4. Identify the answer

The calculation shows that 20 dm³ of carbon(II) oxide is produced. This corresponds to option B.

5. Tip

When dealing with gas stoichiometry, always check the balanced equation first. If all substances involved are gases, you don't need to convert to moles; you can work directly with the volumes using the coefficients from the balanced equation!