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Stoichiometry 18: How many moles of water are produced when 2.5 moles of propane combust with excess oxygen?
Introduction
Have you ever cooked on a gas grill or watched a propane heater in action?
What you witnessed was a real-life combustion reaction taking place.
In chemistry, combustion reactions are more than just dramatic flames—they are balanced chemical processes that allow us to predict the amount of each product formed from a given amount of fuel.
In this article, we’ll walk through how to calculate how many moles of water are produced when 2.5 moles of propane (C₃H₈) combust completely in the presence of excess oxygen.
And by the end, you won’t just have the answer—you’ll understand the reasoning behind every step.
Problem Statement
The question is simple but foundational.
How many moles of water are produced when 2.5 moles of propane (C₃H₈) undergo complete combustion with excess oxygen (O₂)?
The key to solving this is in the balanced chemical equation, which provides the quantitative relationships between all substances involved in the reaction.
Let’s start there.
Step 1: Write the Balanced Chemical Equation
The balanced combustion reaction of propane is:
C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
This equation shows that 1 mole of propane reacts with 5 moles of oxygen to produce 3 moles of carbon dioxide (COâ‚‚) and 4 moles of water (Hâ‚‚O).
These numbers are not arbitrary.
They come from ensuring that the number of each type of atom is conserved on both sides of the equation.
Step 2: Identify the Mole Ratio
To find the moles of water produced, we look at the mole ratio between propane and water in the balanced equation.
From the equation, we see:
1 mole of C₃H₈ → 4 moles of H₂O
That is, every single mole of propane gives rise to four moles of water.
This ratio is the key to solving the problem.
Step 3: Apply the Mole Ratio to the Given Quantity
You’re told that you have 2.5 moles of propane.
Using the mole ratio from Step 2, you multiply:
2.5 moles C₃H₈ × (4 moles H₂O / 1 mole C₃H₈) = 10 moles H₂O
So the answer is:
10 moles of water are produced from the complete combustion of 2.5 moles of propane.
This is your final result, but the logic behind it is just as important as the number itself.
Step 4: What This Ratio Actually Represents
Think of the balanced chemical equation as a recipe.
If one “batch” of combustion starts with 1 mole of propane and yields 4 moles of water, then two and a half batches will yield 2.5 times as much water.
That’s where the 10 moles come from.
You’re simply scaling the reaction up by a factor of 2.5.
The mole ratio gives you a direct mathematical relationship between the reactant and the product.
No additional information is needed, and no molar masses are involved because we’re working entirely in moles.
Step 5: What About the Oxygen?
Notice that the question specifies excess oxygen.
This means that there is more than enough Oâ‚‚ present to ensure that all the propane can react completely.
In stoichiometry, when one reactant is in excess, we focus only on the limiting one—in this case, propane.
So even though the balanced equation includes 5 moles of oxygen per mole of propane, we don’t use that number in our calculation because oxygen is not limiting the reaction.
This example highlights a vital concept in stoichiometry: only the limiting reactant determines how much product forms.
Step 6: Common Errors and How to Avoid Them
Let’s look at some common mistakes students make when solving problems like this.
One mistake is using the coefficient of oxygen instead of the coefficient of water in the mole ratio.
The question asks for moles of water, so your mole ratio must relate propane to water, not propane to oxygen.
Another common mistake occurs when students assume the ratio is 1:1 simply because the question mentions only propane.
It is always the balanced chemical equation that reveals the correct ratio, not the wording of the question alone.
Finally, students sometimes try to convert moles to grams, even though the question is only about moles.
This introduces unnecessary steps and increases the chance of errors.
Staying focused on what the question asks is a crucial exam strategy.
Step 7: Real-World Applications of This Reaction
Propane combustion isn’t just an academic exercise.
Industries and individuals use it extensively in daily activities and manufacturing processes.
Propane fuels gas stoves, barbecues, portable heaters, and even some internal combustion engines.
That’s why knowing how much water forms from a given amount of propane helps engineers manage condensation, calculate energy release, and design effective ventilation systems.
In enclosed environments like greenhouses or industrial ovens, unaccounted-for water vapor can affect humidity levels, pressure, and even chemical stability.
That’s why engineers and chemists must calculate these numbers precisely.
Stoichiometry gives them the confidence to scale reactions safely and efficiently.
Step 8: Visualize the Process
Imagine each propane molecule as a mini fuel pack.
Each one has eight hydrogen atoms that, when burned, combine with oxygen to form water molecules.
Burning 2.5 moles of propane releases enough hydrogen to make 10 moles of water.
This isn’t just theoretical.
In real experiments, this water condenses on cold surfaces, and the amount matches the stoichiometric prediction.
It’s proof that the math works.
Final Wrap-Up
Let’s recap what we learned.
When 2.5 moles of propane combust with excess oxygen, the reaction produces 10 moles of water.
We used the balanced chemical equation for propane combustion to extract the mole ratio: 1 mole of propane produces 4 moles of water.
Then, we applied that ratio to 2.5 moles of propane and found the answer—without needing any information about mass or molar mass.
As a result, this problem shows how stoichiometry connects chemical equations to precise quantitative predictions.
Moreover, it proves that even basic mole conversions can have significant implications in real-world chemistry.
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In addition, you’ll explore video walkthroughs, solve practice problems with explanations, and apply stoichiometry to real-world situations that make the concept click.
Each tutorial helps you understand the concept, rather than just memorizing steps.
Whether you’re prepping for exams or working through chemistry for the first time, CopyChemistry can help you go from confusion to clarity.
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