Gas Stoichiometry Calculator

Calculate volume of gas produced or consumed in reactions using stoichiometry and ideal gas law.

Moles of known reactant/product (mol)

Stoichiometric coefficient of known

Stoichiometric coefficient of target gas

Temperature T (K) (K)

Pressure P (atm) (atm)

Please check your inputs and try again.

The Three-Step Route

Every gas stoichiometry problem follows the same path, and getting lost usually means skipping the middle step.

Step	Operation	Tool
1. To moles	Convert the known quantity to moles	Mass ÷ molar mass, or PV = nRT
2. Mole ratio	Scale by coefficients from the balanced equation	Target coefficient ÷ known coefficient
3. To the answer	Convert target moles to the required unit	PV = nRT for volume

Moles are the currency. Mass cannot be converted to volume directly, and volume ratios only equal mole ratios for gases at identical temperature and pressure — Avogadro's law. That shortcut is genuinely useful when every species is gaseous under the same conditions, letting you skip straight from volume ratio to volume ratio.

Standard Conditions Are Not All the Same

Convention	Temperature	Pressure	Molar volume
STP (IUPAC current)	0°C	1 bar	22.71 L/mol
STP (pre-1982)	0°C	1 atm	22.41 L/mol
SATP	25°C	1 bar	24.79 L/mol
NTP (common industry)	20°C	1 atm	24.06 L/mol

These differ by up to 10%, which matters for any quantitative answer. When a problem says “at STP” without specifying, the 22.4 L/mol convention is usually intended, but the assumption should be stated.

Worked Examples

Example 1: 2H2O2 → 2H2O + O2: 0.5 mol H2O2, find V(O2) at 25°C

mol_O2=0.5×1/2=0.25mol

Result: V=0.25×0.08206×298/1=6.11L

Decomposition of hydrogen peroxide

Example 2: CaCO3 → CaO + CO2: 50g CaCO3, P=1atm, T=400K

n=50/100=0.5mol, c2/c1=1

Result: V=0.5×0.08206×400=16.4L CO2

Industrial calcination reaction

Example 3: Limiting reagent with gases

2H₂ + O₂ → 2H₂O with 3 mol H₂ and 2 mol O₂

Result: H₂ limits; 3 mol H₂O forms

H₂ supports 3 mol of water while O₂ would support 4. The smaller governs, leaving 0.5 mol O₂ unreacted.

Example 4: Volume ratio shortcut

N₂ + 3H₂ → 2NH₃, all gases at the same T and P

Result: 10 L N₂ needs 30 L H₂, gives 20 L NH₃

Volume ratios equal mole ratios for gases under identical conditions, so the coefficients can be applied directly to volumes.

Common Mistakes

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Using volume ratios for non-gaseous species

Avogadro's law applies only to gases at the same temperature and pressure. Solids and liquids must go through moles.

⚠️

Forgetting the mole ratio step

Converting mass to moles then straight to volume without applying the stoichiometric coefficients is the most common error in these problems.

⚠️

Assuming 22.4 L/mol at room temperature

That value is for 0°C. At 25°C the molar volume is nearly 24.5 L/mol — a 9% difference.

⚠️

Not identifying the limiting reagent

When two reactant amounts are given, the smaller mole-adjusted quantity governs. Using the excess reagent overestimates the product.

Frequently Asked Questions

STP vs SATP vs NTP?

STP (IUPAC 2000+): 0°C, 1 bar, molar volume=22.711L. Old STP: 0°C, 1 atm, molar volume=22.414L. SATP: 25°C, 1 bar, molar volume=24.789L. Check which standard is used in your textbook.

Gas stoichiometry at non-standard conditions?

Use PV=nRT with actual T and P. Mole ratio from balanced equation applies regardless of conditions. Don't assume 22.4L/mol unless at exactly 0°C, 1atm.

Can I use volume ratios directly?

Only when all species are gases at the same temperature and pressure. Avogadro's law makes volume proportional to moles under those conditions.

What molar volume should I use?

22.41 L/mol at 0°C and 1 atm, 22.71 at 0°C and 1 bar, or 24.79 at 25°C and 1 bar. State which convention you are using.

How do I find the limiting reagent?

Divide each reactant's moles by its coefficient. The smallest result identifies the limiting reagent and governs the product amount.

Why must I convert to moles first?

Because balanced equations relate moles, not masses or volumes. Mass and volume relate to moles by different conversion factors for each substance.

Formula Explorer connections

Interpretation: This relationship converts chemical amount, mass, composition or balanced-equation ratios into a reaction quantity. Assumption: Use a balanced reaction, consistent units and the correct molar mass. Purity, side reactions and limiting reagents can change experimental results.

Gay-Lussac's Law Calculator →Grams to Moles Calculator →Limiting Reagent Calculator →Chemistry Formula Explorer →