Stoichiometry and Conversion Factors

Oct 13, 2025 by windailys.com 58 views

Understanding the Reaction

The given reaction is $3 O_2 \rightarrow 2 O_3$ . This equation tells us that 3 moles of oxygen gas $(O_2)$ are converted into 2 moles of ozone $(O_3)$ . This is a crucial piece of information when it comes to converting between the two gases using stoichiometry.

Stoichiometry and Conversion Factors

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in chemical reactions. In this case, we are interested in converting from oxygen gas to ozone gas. To do this, we need to use a conversion factor that relates the two gases.

Finding the Correct Conversion Factor

The correct conversion factor can be found by looking at the coefficients in the balanced equation. In this case, the coefficients are 3 for oxygen gas and 2 for ozone gas. This means that for every 3 moles of oxygen gas, we get 2 moles of ozone gas.

Calculating the Conversion Factor

To calculate the conversion factor, we can divide the number of moles of ozone gas by the number of moles of oxygen gas. This gives us:

\frac{2 \text{ moles } O_3}{3 \text{ moles } O_2}

This is our conversion factor. We can use it to convert from oxygen gas to ozone gas.

Converting from Oxygen Gas to Ozone Gas

Now that we have our conversion factor, we can use it to convert from 15 liters of oxygen gas to 10 liters of ozone gas. To do this, we need to know the volume of oxygen gas in moles. We can use the ideal gas law to find this:

PV = nRT

where $P$ is the pressure, $V$ is the volume, $n$ is the number of moles, $R$ is the gas constant, and $T$ is the temperature.

Finding the Number of Moles of Oxygen Gas

We are given that the volume of oxygen gas is 15 liters. We can use the ideal gas law to find the number of moles of oxygen gas:

n = \frac{PV}{RT}

We can plug in the values for $P$ , $V$ , $R$ , and $T$ to find the number of moles of oxygen gas.

Calculating the Number of Moles of Oxygen Gas

Let's assume that the pressure is 1 atm, the volume is 15 liters, the temperature is 298 K, and the gas constant is 0.0821 L atm/mol K. Plugging in these values, we get:

n = \frac{(1 \text{ atm})(15 \text{ L})}{(0.0821 \text{ L atm/mol K})(298 \text{ K})}

Solving for $n$ , we get:

n = 0.6 \text{ moles } O_2

Using the Conversion Factor to Convert to Ozone Gas

Now that we have the number of moles of oxygen gas, we can use our conversion factor to convert to ozone gas:

\frac{2 \text{ moles } O_3}{3 \text{ moles } O_2} \times 0.6 \text{ moles } O_2

Simplifying, we get:

\frac{2}{3} \times 0.6 = 0.4 \text{ moles } O_3

Finding the Volume of Ozone Gas

Now that we have the number of moles of ozone gas, we can use the ideal gas law to find the volume of ozone gas:

V = \frac{nRT}{P}

Plugging in the values for $n$ , $R$ , $T$ , and $P$ , we get:

V = \frac{(0.4 \text{ moles})(0.0821 \text{ L atm/mol K})(298 \text{ K})}{(1 \text{ atm})}

Solving for $V$ , we get:

V = 9.8 \text{ liters } O_3

Conclusion

In this article, we used stoichiometry to convert from 15 liters of oxygen gas to 10 liters of ozone gas. We found the correct conversion factor by looking at the coefficients in the balanced equation and used it to convert from oxygen gas to ozone gas. We also used the ideal gas law to find the volume of ozone gas.

Answer

Q: What is stoichiometry and how is it used in this problem?

A: Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in chemical reactions. In this problem, we used stoichiometry to convert from oxygen gas to ozone gas by using the coefficients in the balanced equation.

Q: What is the balanced equation for the reaction between oxygen and ozone?

A: The balanced equation for the reaction between oxygen and ozone is $3 O_2 \rightarrow 2 O_3$ .

Q: How do we find the correct conversion factor for this reaction?

A: We find the correct conversion factor by looking at the coefficients in the balanced equation. In this case, the coefficients are 3 for oxygen gas and 2 for ozone gas. This means that for every 3 moles of oxygen gas, we get 2 moles of ozone gas.

Q: What is the conversion factor for this reaction?

A: The conversion factor for this reaction is $\frac{2 \text{ moles } O_3}{3 \text{ moles } O_2}$ .

Q: How do we use the conversion factor to convert from oxygen gas to ozone gas?

A: We use the conversion factor by multiplying the number of moles of oxygen gas by the conversion factor. For example, if we have 0.6 moles of oxygen gas, we can convert it to ozone gas by multiplying it by the conversion factor: $\frac{2 \text{ moles } O_3}{3 \text{ moles } O_2} \times 0.6 \text{ moles } O_2$ .

Q: What is the volume of ozone gas produced from 15 liters of oxygen gas?

A: To find the volume of ozone gas produced from 15 liters of oxygen gas, we need to know the number of moles of oxygen gas. We can use the ideal gas law to find this: $n = \frac{PV}{RT}$ . Once we have the number of moles of oxygen gas, we can use the conversion factor to convert it to ozone gas.

Q: What is the ideal gas law and how is it used in this problem?

A: The ideal gas law is $PV = nRT$ , where $P$ is the pressure, $V$ is the volume, $n$ is the number of moles, $R$ is the gas constant, and $T$ is the temperature. We use the ideal gas law to find the number of moles of oxygen gas by rearranging the equation to solve for $n$ : $n = \frac{PV}{RT}$ .

Q: What is the final answer for the volume of ozone gas produced from 15 liters of oxygen gas?

A: The final answer for the volume of ozone gas produced from 15 liters of oxygen gas is 9.8 liters.

Q: What is the correct answer choice for this problem?

A: The correct answer choice for this problem is B. $\frac{15}{2}$ .