Question

Consider the formation of nitrogen dioxide from nitric oxide and oxygen: $$ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) $$ If \(9.0 \mathrm{~L}\) of \(\mathrm{NO}\) is combined with excess \(\mathrm{O}_{2}\) at STP, what is the volume in liters of the \(\mathrm{NO}_{2}\) produced?

Short answer

9.0 L of \(\mathrm{NO}_{2}\) is produced.

Step-by-step solution

Understand the Reaction at STP

The reaction given is \(2 \mathrm{NO}(g) + \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{NO}_{2}(g)\). Under standard temperature and pressure (STP) conditions, 1 mole of any gas occupies 22.4 L. This information will be essential for calculations in the next steps.

Convert Volume of NO to Moles

Using the molar volume at STP, convert the given volume of \(\mathrm{NO}\) to moles. Since \(1\) mole of a gas at STP equals \(22.4\) L, calculate the moles of \(\mathrm{NO}\) as follows:\[\text{moles of NO} = \frac{9.0 \text{ L}}{22.4 \text{ L/mol}} = 0.402 \text{ mol}\]

Use Stoichiometry to Find Moles of NO2

According to the balanced chemical equation, \(2\) moles of \(\mathrm{NO}\) produce \(2\) moles of \(\mathrm{NO}_{2}\). Therefore, the moles of \(\mathrm{NO}_{2}\) produced will be the same as the moles of \(\mathrm{NO}\). So, we have:\[0.402 \text{ mol of NO} \rightarrow 0.402 \text{ mol of NO}_{2}\]

Convert Moles of NO2 to Volume

Convert the moles of \(\mathrm{NO}_{2}\) back to volume using the molar volume at STP (22.4 L/mole):\[\text{Volume of NO}_{2} = 0.402 \text{ mol} \times 22.4 \text{ L/mol} = 9.0 \text{ L}\]

Conclusion

Since we performed the calculations correctly, the volume of \(\mathrm{NO}_{2}\) produced is equal to the initial volume of \(\mathrm{NO}\). Therefore, \(9.0\) L of \(\mathrm{NO}_{2}\) is produced.

Key concepts

Chemical Reactions

A chemical reaction involves the transformation of reactants into products. In our example, the reaction is between nitric oxide (NO) and oxygen (O extsubscript{2}), which forms nitrogen dioxide (NO extsubscript{2}). The balanced chemical equation is crucial to understanding the stoichiometry—the quantitative relationship between reactants and products. Here, the equation is \( 2 \text{NO}(g) + \text{O}_2(g) \rightarrow 2 \text{NO}_2(g) \). This tells us:

• 2 moles of NO combine with 1 mole of O extsubscript{2} to produce 2 moles of NO extsubscript{2}.
• The coefficients in the equation represent the ratio in which the chemicals react.

Understanding these ratios is critical for volume and mole calculations, which you'll use to determine how much product (NO extsubscript{2}) is formed.

Gas Laws

Gas laws describe how gases behave based on factors like volume, temperature, and pressure. A key concept is that the behavior of gases is predictable and follows certain principles:

• At a constant temperature and pressure, the volume of a gas is directly proportional to the number of moles. This is known as Avogadro's Law.
• Under standard temperature and pressure conditions (STP)—0°C (273 K) and 1 atm pressure—1 mole of any ideal gas occupies a volume of 22.4 liters.

These principles are crucial for understanding the stoichiometric calculations that involve gases.

Mole Concept

The mole is a fundamental concept in chemistry that relates quantities on a macroscopic level to numbers of atoms or molecules. In this exercise, it's used to relate the mass of a substance to the volume of gases in a reaction:

• A mole represents \( 6.022 \times 10^{23} \) entities (atoms, molecules, etc.).
• Using the mole concept, we convert the volume of NO gas to moles using the molar volume at STP (22.4 L/mol).

This conversion helps in making predictions about the volume of product formed in the reaction, such as NO extsubscript{2} in our case.

STP

Standard Temperature and Pressure (STP) is a reference point used in chemistry to describe the conditions of a gas sample:

• STP is defined as 0°C (273 K) and 1 atm pressure.
• These conditions allow us to use 22.4 liters as the volume of 1 mole of an ideal gas.

In this exercise, STP conditions simplify our calculations, allowing us to directly relate volumes and moles using the molar volume concept.

Volume Calculations

Volume calculations in stoichiometry involve using known values from a balanced chemical equation to find unknown quantities. We used this for the reaction:

• First, we calculated the moles of NO given the volume (9.0 L) using the molar volume at STP (22.4 L/mol).
• Next, we used stoichiometry to find moles of NO extsubscript{2}, matching the moles of NO.
• Finally, we converted moles of NO extsubscript{2} back to volume, arriving again at 9.0 L.

This shows the conservation of volume under these reactions and conditions, emphasizing the importance of stoichiometry in predicting chemical reactions.