Propane Combustion: Calculate Oxygen Needed For CO2 Production

Propane $\left( C _3 H _8\right)$ is a widely used fuel, powering everything from backyard grills to some homes. Its combustion, or reaction with oxygen, is a fundamental chemical process that yields carbon dioxide and water. Understanding the stoichiometry of this reaction is crucial for various applications, especially when calculating the exact amounts of reactants needed or products formed. The balanced chemical equation that governs this process is:

$C_3 H_8 + 5 O_2 \rightarrow 3 CO_2 + 4 H_2 O$

This equation tells us that for every one molecule of propane that burns, five molecules of oxygen are consumed, producing three molecules of carbon dioxide and four molecules of water. In this article, we'll delve into how to use this equation to answer a specific question: how many grams of oxygen are required to produce $37.15 g CO _2$? This involves a step-by-step calculation using molar masses and the mole ratios derived from the balanced equation. It's a great way to see practical chemistry in action!

Understanding Stoichiometry and the Balanced Equation

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in chemical reactions. It's essentially the 'accounting' of atoms and molecules in a chemical change. The balanced chemical equation, $C_3 H_8 + 5 O_2 \rightarrow 3 CO_2 + 4 H_2 O$, is our foundational tool. It's not just a statement of what reacts and what forms; it's a precise ratio. The coefficients in front of each chemical formula represent the relative number of moles of each substance involved. In this specific reaction, the coefficients tell us:

• 1 mole of propane reacts with 5 moles of oxygen.
• This reaction produces 3 moles of carbon dioxide and 4 moles of water.

The key mole ratio we're interested in for this problem is the one between carbon dioxide ($CO_2$) and oxygen ($O_2$). From the equation, we see that 3 moles of $CO_2$ are produced for every 5 moles of $O_2$ consumed. This ratio, $5 ext{ moles } O_2 / 3 ext{ moles } CO_2$, is fundamental to solving our problem. It allows us to bridge the gap between the amount of carbon dioxide we want to produce and the amount of oxygen needed to achieve it.

Before we can use this mole ratio, we need to convert the given mass of carbon dioxide into moles. This requires knowing the molar mass of $CO_2$. The molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol). We calculate it by summing the atomic masses of all atoms in the molecule. For $CO_2$:

• Carbon (C) has an atomic mass of approximately 12.01 g/mol.
• Oxygen (O) has an atomic mass of approximately 16.00 g/mol.

So, the molar mass of $CO_2$ is $(1 imes 12.01 ext{ g/mol}) + (2 imes 16.00 ext{ g/mol}) = 12.01 + 32.00 = 44.01 ext{ g/mol}$.

Similarly, we'll need the molar mass of oxygen ($O_2$) for the final step. The atomic mass of oxygen is approximately 16.00 g/mol. Since oxygen exists as a diatomic molecule ($O_2$), its molar mass is $2 imes 16.00 ext{ g/mol} = 32.00 ext{ g/mol}$.

These molar masses are essential conversion factors that allow us to move between the macroscopic world of grams and the microscopic world of moles, which is where the balanced equation's ratios operate. With these tools in hand, we are ready to tackle the calculation.

Step-by-Step Calculation: From Grams of CO2 to Grams of O2

Now, let's put our knowledge of stoichiometry and molar masses to work to answer the question: How many grams of oxygen are required to produce $37.15 g CO _2$? We will break this down into manageable steps.

Step 1: Convert the mass of $CO_2$ to moles of $CO_2$.

We are given $37.15 g$ of $CO_2$. To convert this mass to moles, we use the molar mass of $CO_2$ (which we calculated as $44.01 ext{ g/mol}$) as a conversion factor:

ext{Moles of } CO_2 = rac{ ext{Mass of } CO_2}{ ext{Molar Mass of } CO_2} = rac{37.15 ext{ g}}{44.01 ext{ g/mol}}

Performing this calculation:

$$ext{Moles of } CO_2 \approx 0.8441 ext{ mol}$$

So, we have approximately $0.8441$ moles of carbon dioxide. This is the amount we want to produce.

Step 2: Use the mole ratio from the balanced equation to find moles of $O_2$.

Our balanced equation, $C_3 H_8 + 5 O_2 \rightarrow 3 CO_2 + 4 H_2 O$, shows that $5$ moles of $O_2$ are required to produce $3$ moles of $CO_2$. This gives us the mole ratio: $\frac{5 ext{ moles } O_2}{3 ext{ moles } CO_2}$. We multiply the moles of $CO_2$ we just calculated by this ratio to find out how many moles of $O_2$ are needed:

ext{Moles of } O_2 = ( ext{Moles of } CO_2) \times rac{5 ext{ moles } O_2}{3 ext{ moles } CO_2}

Substituting the value from Step 1:

ext{Moles of } O_2 = 0.8441 ext{ mol } CO_2 \times rac{5 ext{ moles } O_2}{3 ext{ moles } CO_2}

Notice how the 'moles of $CO_2$' units cancel out, leaving us with 'moles of $O_2$':

$$ext{Moles of } O_2 \approx 0.8441 \times 1.6667 ext{ mol } O_2 \approx 1.4068 ext{ mol } O_2$$

So, we need approximately $1.4068$ moles of oxygen gas.

Step 3: Convert moles of $O_2$ to grams of $O_2$.

Finally, we convert the moles of oxygen back into grams using its molar mass (which we found to be $32.00 ext{ g/mol}$):

$$ext{Mass of } O_2 = ( ext{Moles of } O_2) \times ( ext{Molar Mass of } O_2)$$

Plugging in our values:

$$ext{Mass of } O_2 = 1.4068 ext{ mol } O_2 \times 32.00 ext{ g/mol}$$

Calculating the final mass:

$$ext{Mass of } O_2 \approx 45.018 ext{ g}$$

Therefore, approximately $45.02$ grams of oxygen are required to produce $37.15$ grams of carbon dioxide during the combustion of propane.

Practical Implications and Further Considerations

This calculation demonstrates a core principle in chemistry: the ability to predict and control the amounts of substances involved in a chemical reaction. Understanding these quantitative relationships has significant practical implications. For instance, in industrial processes involving combustion, precise control over reactant ratios is essential for efficiency, safety, and minimizing unwanted byproducts. If a process requires a certain amount of heat output, which is directly related to the amount of fuel burned, then calculating the necessary amount of oxygen (or air, which is a mixture containing oxygen) is paramount.

In the context of propane, this could apply to:

• Engine Performance: Ensuring optimal fuel-air mixtures in internal combustion engines that run on propane to maximize power and fuel economy.
• Industrial Furnaces and Boilers: Maintaining efficient combustion for heating applications, preventing incomplete combustion which can lead to soot formation and reduced energy output.
• Safety Systems: In environments where propane is used, understanding its combustion characteristics helps in designing ventilation and safety measures.

It's also important to remember that real-world conditions are often more complex than a simple balanced equation. Air, for example, is not pure oxygen but a mixture, primarily nitrogen (about 78%), oxygen (about 21%), and small amounts of other gases. When calculating the amount of air needed, one must account for this composition. The presence of nitrogen, while largely inert in the combustion process itself, can affect heat transfer and the overall volume of gases involved.

Furthermore, factors like temperature, pressure, and the presence of catalysts can influence reaction rates and even the products formed. However, the stoichiometric calculation provides a fundamental baseline – the theoretical ideal. Deviations from this ideal are often what chemists and engineers study to optimize processes.

For those interested in the broader context of chemical reactions and their applications, exploring resources on chemical kinetics (the study of reaction rates) and thermodynamics (the study of energy changes in reactions) can provide deeper insights. Understanding the energy released during propane combustion, for example, is crucial for calculating its heating value.

This specific problem, while seemingly straightforward, is a gateway to understanding more complex chemical engineering and industrial chemistry principles. It highlights how fundamental chemical laws are applied to solve real-world problems, ensuring that we can harness chemical reactions safely and efficiently.

Conclusion

In summary, by utilizing the balanced chemical equation for propane combustion ($C_3 H_8 + 5 O_2 \rightarrow 3 CO_2 + 4 H_2 O$) and applying principles of stoichiometry, we determined that approximately $45.02$ grams of oxygen ($O_2$) are required to produce $37.15$ grams of carbon dioxide ($CO_2$). This calculation involved converting the mass of $CO_2$ to moles, using the mole ratio from the balanced equation to find the moles of $O_2$ needed, and then converting those moles back into grams. This process is a cornerstone of quantitative chemistry, enabling us to precisely manage chemical reactions for a vast array of applications, from laboratory experiments to large-scale industrial processes.

For further reading on the fundamental principles of chemistry, including stoichiometry and chemical reactions, I recommend exploring resources from reputable scientific organizations. A great starting point for understanding chemical concepts and their real-world applications is the American Chemical Society website. You can find valuable information on their educational resources and publications that delve deeper into topics like combustion and chemical calculations.