The best examples of ideal gas law in stoichiometry: 3 examples (plus more)

Before we talk theory, it helps to anchor everything in concrete problems. When teachers talk about examples of ideal gas law in stoichiometry: 3 examples, they usually mean three classic scenarios:

• A gas as a product whose mass or moles you want to find
• A gas as a reactant supplied at some pressure, volume, and temperature
• A gas mixture where stoichiometry and the ideal gas law work together

We’ll walk through those three in depth, then add several more real examples so you can see how the same logic repeats.

Example 1: Hydrogen from magnesium and acid (gas as a product)

Reaction:

Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

Scenario: A strip of magnesium reacts with excess hydrochloric acid. The hydrogen gas is collected over water at 25 °C in a 2.50 L container at 0.980 atm total pressure. How many grams of Mg reacted?

This is a textbook example of the ideal gas law feeding into stoichiometry.

Step 1 – Use PV = nRT to get moles of H₂.
Take R = 0.08206 L·atm·mol⁻¹·K⁻¹, T = 25 °C = 298 K.

[
P = 0.980\,\text{atm},\; V = 2.50\,\text{L},\; T = 298\,\text{K}
]

[
n_{H_2} = \frac{PV}{RT} = \frac{0.980 \times 2.50}{0.08206 \times 298} \approx 0.100\,\text{mol H}_2
]

(If you’re collecting over water in a real lab, you’d subtract water vapor pressure; the NIST Chemistry WebBook lists vapor pressure values.)

Step 2 – Use stoichiometry to relate H₂ to Mg.
From the balanced equation, 1 mol Mg → 1 mol H₂.

[
0.100\,\text{mol H}_2 \Rightarrow 0.100\,\text{mol Mg}
]

Step 3 – Convert moles of Mg to mass.
Molar mass of Mg ≈ 24.3 g/mol.

[
0.100\,\text{mol Mg} \times 24.3\,\text{g/mol} = 2.43\,\text{g Mg}
]

This first entry in our examples of ideal gas law in stoichiometry: 3 examples shows the core pattern: ideal gas law → moles of gas → stoichiometric ratio → mass.

Example 2: Ammonia synthesis (gas as a reactant at non‑STP conditions)

Reaction (Haber process):

N₂(g) + 3H₂(g) → 2NH₃(g)

Scenario: You have a tank containing hydrogen at 5.00 atm and 350 K with a volume of 10.0 L. Nitrogen is in excess. How many grams of NH₃ can you theoretically produce from the hydrogen in the tank?

This is a classic industrial-style example of ideal gas law in stoichiometry because it mirrors how real ammonia plants plan feedstock consumption.

Step 1 – Use PV = nRT for H₂.

[
P = 5.00\,\text{atm},\; V = 10.0\,\text{L},\; T = 350\,\text{K}
]

[
n_{H_2} = \frac{PV}{RT} = \frac{5.00 \times 10.0}{0.08206 \times 350} \approx 1.73\,\text{mol H}_2
]

Step 2 – Use stoichiometry H₂ → NH₃.
From the balanced equation, 3 mol H₂ → 2 mol NH₃.

[
1.73\,\text{mol H}_2 \times \frac{2\,\text{mol NH}_3}{3\,\text{mol H}_2} \approx 1.15\,\text{mol NH}_3
]

Step 3 – Convert moles of NH₃ to grams.
Molar mass NH₃ ≈ 17.0 g/mol.

[
1.15\,\text{mol NH}_3 \times 17.0\,\text{g/mol} \approx 19.6\,\text{g NH}_3
]

This second entry in our examples of ideal gas law in stoichiometry: 3 examples highlights a common exam theme: a gas reactant at arbitrary conditions, not STP.

For context, modern ammonia production is monitored at the scale of millions of tons per year; the U.S. Energy Information Administration (EIA) tracks natural gas use that feeds into hydrogen for the Haber process. The same PV = nRT logic is used in process calculations, just with bigger numbers and real‑gas corrections.

Example 3: Combustion of propane (gas mixture and limiting reactant)

Reaction:

C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(g)

Scenario: A grill cylinder contains 2.00 L of propane at 1.20 atm and 298 K. It’s burned in a chamber that initially contains 10.0 L of O₂ at 1.00 atm and 298 K. Which gas is limiting, and how many liters of CO₂ will form at 298 K and 1.00 atm?

This is a richer example of the ideal gas law in stoichiometry because you have gas reactants on both sides, a limiting reactant, and a gas product.

Step 1 – Moles of propane and oxygen using PV = nRT.

Propane:
[
P = 1.20\,\text{atm},\; V = 2.00\,\text{L},\; T = 298\,\text{K}
]
[
n_{C_3H_8} = \frac{1.20 \times 2.00}{0.08206 \times 298} \approx 0.098\,\text{mol}
]

Oxygen:
[
P = 1.00\,\text{atm},\; V = 10.0\,\text{L},\; T = 298\,\text{K}
]
[
n_{O_2} = \frac{1.00 \times 10.0}{0.08206 \times 298} \approx 0.409\,\text{mol}
]

Step 2 – Determine the limiting reactant.
The stoichiometric ratio is 1 mol C₃H₈ : 5 mol O₂.

Required O₂ for 0.098 mol C₃H₈:
[
0.098 \times 5 = 0.490\,\text{mol O}_2
]

You only have 0.409 mol O₂, so O₂ is limiting.

Step 3 – Use limiting reactant to find moles of CO₂.
From the equation, 5 mol O₂ → 3 mol CO₂.

[
0.409\,\text{mol O}_2 \times \frac{3\,\text{mol CO}_2}{5\,\text{mol O}_2} \approx 0.245\,\text{mol CO}_2
]

Step 4 – Convert moles of CO₂ to volume via PV = nRT.
At 1.00 atm and 298 K:

[
V_{CO_2} = \frac{nRT}{P} = \frac{0.245 \times 0.08206 \times 298}{1.00} \approx 5.99\,\text{L CO}_2
]

So from this third entry in our examples of ideal gas law in stoichiometry: 3 examples, you get about 6.0 L of CO₂ under the stated conditions.

More real examples of ideal gas law in stoichiometry (beyond the “3 examples”)

Those three are the “greatest hits” your teacher expects, but exam writers and lab instructors love to remix them. Here are several more real examples of ideal gas law in stoichiometry that show up in classrooms, labs, and research.

Decomposition of sodium azide in airbags

Reaction:

2NaN₃(s) → 2Na(s) + 3N₂(g)

Scenario: Older airbag designs used sodium azide. Suppose 65.0 g of NaN₃ are triggered, and the nitrogen gas inflates a 36.0 L bag at 1.10 atm and 298 K. Is that mass of NaN₃ enough to fill the bag?

Step 1 – Moles of N₂ needed from PV = nRT.

[
n_{N_2} = \frac{PV}{RT} = \frac{1.10 \times 36.0}{0.08206 \times 298} \approx 1.62\,\text{mol N}_2
]

Step 2 – Moles of NaN₃ required from stoichiometry.
From the equation, 2 mol NaN₃ → 3 mol N₂.

[
1.62\,\text{mol N}_2 \times \frac{2\,\text{mol NaN}_3}{3\,\text{mol N}_2} \approx 1.08\,\text{mol NaN}_3
]

Molar mass NaN₃ ≈ 65.0 g/mol.

Required mass:
[
1.08\,\text{mol} \times 65.0\,\text{g/mol} \approx 70.2\,\text{g NaN}_3
]

Since 65.0 g < 70.2 g, that charge of NaN₃ is not quite enough to reach those conditions. This is one of the best examples of how the ideal gas law in stoichiometry connects directly to safety engineering.

For a deeper look at gas behavior and safety standards in vehicles, the U.S. National Highway Traffic Safety Administration (NHTSA) provides technical reports and testing data.

Laboratory oxygen from potassium chlorate

Reaction:

2KClO₃(s) → 2KCl(s) + 3O₂(g)

Scenario: In a lab, you heat KClO₃ to generate oxygen. You collect 750 mL of O₂ at 0.960 atm and 305 K. How many grams of KClO₃ decomposed?

Step 1 – PV = nRT for O₂.

[
P = 0.960\,\text{atm},\; V = 0.750\,\text{L},\; T = 305\,\text{K}
]

[
n_{O_2} = \frac{0.960 \times 0.750}{0.08206 \times 305} \approx 0.0299\,\text{mol O}_2
]

Step 2 – Stoichiometry to find moles of KClO₃.
2 mol KClO₃ → 3 mol O₂.

[
0.0299\,\text{mol O}_2 \times \frac{2\,\text{mol KClO}_3}{3\,\text{mol O}_2} \approx 0.0199\,\text{mol KClO}_3
]

Step 3 – Convert to mass.
Molar mass KClO₃ ≈ 122.6 g/mol.

[
0.0199\,\text{mol} \times 122.6\,\text{g/mol} \approx 2.44\,\text{g KClO}_3
]

This is a lab‑friendly example of ideal gas law in stoichiometry that you can actually run in a high school setting (with proper safety protocols).

CO₂ production from yeast fermentation

Simplified reaction for glucose fermentation:

C₆H₁₂O₆(aq) → 2C₂H₅OH(aq) + 2CO₂(g)

Scenario: A fermentation setup produces 4.50 L of CO₂ at 1.05 atm and 298 K. How many grams of glucose were consumed?

Step 1 – PV = nRT for CO₂.

[
n_{CO_2} = \frac{1.05 \times 4.50}{0.08206 \times 298} \approx 0.193\,\text{mol CO}_2
]

Step 2 – Stoichiometry CO₂ → glucose.
1 mol glucose → 2 mol CO₂.

[
0.193\,\text{mol CO}_2 \times \frac{1\,\text{mol glucose}}{2\,\text{mol CO}_2} \approx 0.0965\,\text{mol glucose}
]

Molar mass glucose ≈ 180.2 g/mol.

[
0.0965\,\text{mol} \times 180.2\,\text{g/mol} \approx 17.4\,\text{g glucose}
]

Fermentation is one of the best real examples of ideal gas law in stoichiometry connecting chemistry to food science and brewing.

Atmospheric chemistry: ozone formation and NO₂

Ozone chemistry in the troposphere involves nitrogen oxides and oxygen. A simplified formation step:

NO₂(g) + O₂(g) → NO(g) + O₃(g)

Scenario: In a controlled smog‑chamber experiment at 298 K, 2.00 L of NO₂ at 0.750 atm reacts completely with excess O₂. What volume of O₃ is formed at 298 K and 1.00 atm?

Step 1 – PV = nRT for NO₂.

[
n_{NO_2} = \frac{0.750 \times 2.00}{0.08206 \times 298} \approx 0.0612\,\text{mol NO}_2
]

Step 2 – Stoichiometry NO₂ → O₃.
From the equation, 1 mol NO₂ → 1 mol O₃.

[
n_{O_3} = 0.0612\,\text{mol}
]

Step 3 – Use PV = nRT to find O₃ volume.

[
V_{O_3} = \frac{nRT}{P} = \frac{0.0612 \times 0.08206 \times 298}{1.00} \approx 1.50\,\text{L O}_3
]

This is a clean example of ideal gas law in stoichiometry that mirrors how atmospheric chemists interpret smog‑chamber and air‑quality data. For broader context on ozone and NO₂, NASA’s Earth Observatory and the U.S. EPA’s air quality pages (EPA Air Topics) provide data and visualizations updated through 2024.

Why these are the best examples of ideal gas law in stoichiometry

Across all of these problems, the pattern is the same, even though the chemistry changes:

• A gas variable (P, V, T) is given instead of moles.
• You apply PV = nRT to convert gas conditions into moles.
• Stoichiometry then links those moles to another species: a solid, liquid, or another gas.
• You might convert back to mass or volume at new conditions.

So when teachers talk about examples of ideal gas law in stoichiometry: 3 examples, they’re really training you to see that PV = nRT is just another way to say, “Here’s how many moles of this gas you actually have.”

In real practice—whether it’s ammonia production, airbag deployment, or atmospheric modeling—chemists and engineers still start from the same relationship. At higher pressures and very low temperatures, they switch to real‑gas equations, but the stoichiometric logic doesn’t change.

For more background on the ideal gas law and its limits, the Chemistry LibreTexts project (ChemLibreTexts) and many university general chemistry sites (for example, MIT OpenCourseWare) provide detailed notes and problem sets updated regularly.

FAQ: common questions about examples of ideal gas law in stoichiometry

Q1. What are some common exam examples of ideal gas law in stoichiometry?
Common exam examples of the ideal gas law in stoichiometry include:

• Metal + acid reactions producing H₂(g), where you measure gas volume and back‑calculate metal mass.
• Combustion of hydrocarbons (like propane) with O₂(g) where you track limiting reactants and CO₂ volume.
• Decomposition reactions (NaN₃, KClO₃) where a solid produces a gas that you measure.

Q2. How do I know when to use the ideal gas law in a stoichiometry problem?
Look for any gas with given P, V, T instead of moles or grams. That’s your hint to use PV = nRT to convert conditions to moles before doing any mole ratios. If a problem gives gas volume specifically at STP, you can use molar volume (22.4 L/mol) instead, but PV = nRT always works.

Q3. Can I mix the ideal gas law with limiting reactant problems?
Yes, and many of the best examples of ideal gas law in stoichiometry do exactly that. You convert each gas reactant’s conditions to moles, compare required vs. available moles using the balanced equation, identify the limiting reactant, and then use that to predict product amounts.

Q4. Are these examples valid in real industry, or only in the classroom?
They’re simplified, but the structure is the same in real processes. Industrial systems account for non‑ideal behavior, heat transfer, and safety margins, yet they still rely on stoichiometric relationships and gas‑law‑type calculations. Organizations like the U.S. Department of Energy and the EIA publish data showing how these calculations scale up to power plants and chemical facilities.

Q5. Where can I find more worked examples and practice problems?
University general chemistry sites and open textbooks are great sources. Chemistry LibreTexts and MIT OpenCourseWare both offer multiple examples of ideal gas law in stoichiometry with step‑by‑step solutions and practice sets you can download for free.