Module 2 · L15 of 2040 min⚡ +50 XP in Learn · +25 to completeInquiry Question 4 · Gas Laws
Gas Stoichiometry
Gas stoichiometry is not a new method, it is the 4-step method with one extra conversion step added. When a gas is given or asked for, you convert between volume and moles using molar volume, then proceed as normal. The only trap is choosing the right molar volume for the stated conditions.
Today's hook, Gas stoichiometry is not a new method, it is the 4-step method with one extra conversion step added. When a gas is given or asked for, you convert between volume and moles using molar volume, then proceed as normal. The only trap is choosing the right molar volume for the stated conditions.
0/5QUESTS
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You're here
Warm up and think first
Warm up first
Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.
Worksheets
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.
Imagine you burn a piece of magnesium ribbon in oxygen: 2Mg + O₂ → 2MgO. If you know the mass of magnesium used, what extra step would you need to find the volume of oxygen gas consumed, and why can't you just use the same 4-step method you already know?
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The gas formulas, and the words for them
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Formula reference · this lesson
core formula
📐
Gas Stoichiometry Formulas
n = V ÷ molar volume (gas volume → moles)
V = n × molar volume (moles → gas volume)
Gas laws (T in kelvin, T(K) = T(°C) + 273.15):
Boyle: P₁V₁ = P₂V₂ | Charles: V₁/T₁ = V₂/T₂ | Gay-Lussac: P₁/T₁ = P₂/T₂
Combined: P₁V₁/T₁ = P₂V₂/T₂
Ideal gas law: PV = nRT, R = 8.314 (kPa·L·mol⁻¹·K⁻¹)
⚠️ The most common error in this lesson is using 22.71 L/mol when the question specifies room temperature or RTP, or using 24.8 L/mol when the question specifies STP or standard conditions. Always identify conditions before choosing the molar volume value.
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What you'll master
Know
Key facts
STP = 0°C, 100 kPa → 22.71 L/mol (NESA standard)
RTP = 25°C, 100 kPa → 24.8 L/mol
Temperature in gas laws must be in kelvin: T(K) = T(°C) + 273.15
Ideal gas law: PV = nRT, R = 8.314 (kPa·L·mol⁻¹·K⁻¹)
Understand
Concepts
Boyle, Charles, Gay-Lussac and Avogadro each hold two variables constant
The combined and ideal gas laws unify them into one relationship
Molar volume only applies at one fixed set of conditions
Gas stoichiometry links gas volumes to moles via the gas laws
Can do
Skills
Convert between °C and K and apply P₁V₁/T₁ = P₂V₂/T₂
Solve for any one of P, V, n or T using PV = nRT
Find a gas volume produced or consumed in a reaction at any conditions
Read and interpret P–V, V–T and P–T graphs
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Key terms
Molar volume (Vₘ)
The volume occupied by one mole of any ideal gas; 22.71 L mol⁻¹ at STP (0°C, 100 kPa); 24.8 L mol⁻¹ at SATP/RTP (25°C, 100 kPa).
STP (Standard Temperature and Pressure)
0°C (273 K) and 100 kPa; molar volume = 22.71 L mol⁻¹ (NESA standard).
RTP (Room Temperature and Pressure)
25°C (298 K) and 100 kPa; molar volume = 24.8 L mol⁻¹.
n = V ÷ Vₘ
Moles of gas = volume ÷ molar volume; applies only when temperature and pressure conditions match the given Vₘ.
Gas stoichiometry calculation
Convert gas volume to moles using n = V/Vₘ; apply mole ratio from balanced equation; convert product moles to mass or volume.
Avogadro's law
Equal volumes of all gases at the same temperature and pressure contain the same number of particles (entities); mole ratios equal volume ratios for gases. V ∝ n at constant T and P.
Boyle's law
At constant temperature and amount, pressure and volume are inversely proportional: P₁V₁ = P₂V₂.
Charles's law
At constant pressure and amount, volume is directly proportional to absolute temperature: V₁/T₁ = V₂/T₂ (T in K).
Gay-Lussac's law
At constant volume and amount, pressure is directly proportional to absolute temperature: P₁/T₁ = P₂/T₂ (T in K).
Ideal gas law
PV = nRT, with R = 8.314 J mol⁻¹ K⁻¹ = 8.314 kPa·L·mol⁻¹·K⁻¹. Relates P, V, n and T for an ideal gas.
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Avogadro's law and molar volume
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Avogadro's Law and Molar Volume
core concept
Avogadro's law states that equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules. This means 1 mole of any gas, regardless of what it is, occupies the same volume under the same conditions.
STP, Standard Temperature and Pressure
22.71 L/mol
Temperature: 0°C (273 K)
Pressure: 100 kPa
Used when question says "STP" or "0°C, 100 kPa"
Current NESA standard. Note: 22.4 L/mol is the older value at 0°C and 1 atm (101.325 kPa), you may see it in older resources. NESA uses 22.71 L/mol at 0°C and 100 kPa.
RTP, Room Temperature and Pressure
24.8 L/mol
Temperature: 25°C (298 K)
Pressure: 100 kPa
Used when question says "RTP", "room conditions", or "25°C"
More realistic for laboratory experiments
The Modified Pathway for Gas Problems
The 4-step stoichiometry method gains one extra step when a gas is involved. If gas volume is given, add a step before Step 2 to convert V → n. If gas volume is the answer, add a step after Step 3 to convert n → V.
Interactive: Gas Volume Calculator
Which extra step applies?
If a gas volume is the input (given) → use Step 0: n = V ÷ MV, then proceed to Step 3 directly.
If a gas volume is the output (asked for) → complete Steps 1–3 normally, then use Step 5: V = n × MV.
If both input and output are gases → use Step 0 AND Step 5.
Avogadro's law: equal volumes of all gases at the same T and P contain equal numbers of molecules. Molar volume: STP (0 °C, 100 kPa) = 22.71 L mol⁻¹; RTP (25 °C, 100 kPa) = 24.8 L mol⁻¹. To find moles from gas volume: n = V ÷ Vm. Always identify conditions before choosing a Vm value.
Pause, copy the highlighted rule and values into your book before moving on.
Odd one out: three of these conditions all use 24.8 L/mol as the molar volume. Which one doesn't belong?
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The gas laws, pressure, volume and temperature
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The gas laws · pressure, volume and temperature
core concept · +5 XP
Molar volume only works at one fixed set of conditions. The gas laws let you predict what a gas does when pressure, volume or temperature change. Each simple law holds two of the four quantities (P, V, T, n) constant and links the other two.
Always use kelvin. Temperature in every gas law must be absolute: T(K) = T(°C) + 273.15. Using °C gives wrong answers (and can divide by zero or go negative). Pressure and volume can be in any consistent units for the ratio laws; for PV = nRT use kPa, L and K.
BBoyle's law
Constant T, n
P and V inversely proportional
P₁V₁ = P₂V₂
CCharles's law
Constant P, n
V directly proportional to T
V₁/T₁ = V₂/T₂
GGay-Lussac's law
Constant V, n
P directly proportional to T
P₁/T₁ = P₂/T₂
AAvogadro's law
Constant P, T
V directly proportional to n
V₁/n₁ = V₂/n₂
Combine the three P–V–T laws into one relationship for a fixed amount of gas:
P₁V₁ / T₁ = P₂V₂ / T₂
Bring in the amount of gas and you get the ideal gas law:
P V = n R T
where R = 8.314 J mol⁻¹ K⁻¹. Because 1 J = 1 kPa·L, you can use R = 8.314 with P in kPa, V in L, T in K directly. The molar volume you already use (22.71 L/mol at STP) is just PV = nRT solved at 0 °C and 100 kPa.
Ideal gas assumptions:
The gas laws assume particles have negligible volume, experience no intermolecular forces, and collide elastically. Real gases deviate most at high pressure and low temperature, where these assumptions break down.
Why kelvin, and why 0 K?
Charles's and Gay-Lussac's graphs above both extrapolate back to the same point, 0 K (−273.15 °C), even though they plot different variables. Kinetic theory explains why: temperature is a measure of the average kinetic energy of a gas's particles, so cooling a gas slows its particles down. Absolute zero is the theoretical temperature at which particle motion, and so kinetic energy, would reach a minimum, which is why the pressure and volume lines above both extrapolate to zero there. It is never actually reached in practice, and no temperature exists below it. That is the reason the kelvin scale has no negative values and every gas law must use it: a °C value can be negative, but a kelvin value describing "less motion than absolute zero" is physically meaningless.
Gas laws (T always in kelvin, T(K) = T(°C) + 273.15): Boyle P₁V₁ = P₂V₂ (constant T); Charles V₁/T₁ = V₂/T₂ (constant P); Gay-Lussac P₁/T₁ = P₂/T₂ (constant V); Avogadro V ∝ n. Combined: P₁V₁/T₁ = P₂V₂/T₂. Ideal gas law: PV = nRT, R = 8.314 (kPa·L·mol⁻¹·K⁻¹). Molar volume is PV = nRT evaluated at one fixed set of conditions.
Pause, copy the five gas-law equations into your book before moving on.
Quick check: A sealed, rigid steel cylinder of gas is heated. Which gas law predicts how its pressure changes, and what happens?
A rigid cylinder fixes the volume, so at constant V and n, Gay-Lussac's law (P₁/T₁ = P₂/T₂) applies: heating raises the absolute temperature, so the pressure rises.
Worked Example, Combined gas law+5 XP on full reveal
A weather balloon holds 15.0 L of helium at ground level (100 kPa, 27 °C). It rises until the pressure is 40.0 kPa and the temperature is −23 °C. Find the new volume.
We just saw the mathematical form of Boyle's, Charles's and Gay-Lussac's laws. That raises a question: how would you actually collect real data to prove P, V and T are related this way? This card answers it → two simple gas-syringe investigations.
Investigating Boyle's law
Aim
To investigate the relationship between pressure and volume of a fixed mass of gas at constant temperature.
Method: Connect a gas syringe, trapping a fixed volume of air, to a pressure sensor or manometer through a short length of tubing, sealed so no gas can escape. Starting with the plunger at a known volume, increase the pressure in small, controlled steps (for example by adding masses to a plunger platform, or using a screw clamp), recording the pressure and the corresponding volume from the syringe scale at each step. Keep the apparatus at constant room temperature throughout, do not touch or warm the syringe barrel, and repeat for a wide range of volumes.
Analysing the data: Plot pressure (P) against volume (V) and a curve results, since P is inversely proportional to V. Plot P against 1/V instead, and a straight line through the origin confirms Boyle's law (P₁V₁ = P₂V₂), with the gradient of that line equal to the constant nRT for the trapped gas.
Investigating Charles's law
Aim
To investigate the relationship between volume and temperature of a fixed mass of gas at constant pressure.
Method: Trap a fixed volume of air in a gas syringe with the plunger free to move, so the pressure stays at atmospheric throughout. Submerge the syringe barrel in a water bath fitted with a thermometer, and heat the bath slowly and evenly using a hotplate or a Bunsen burner and stand. Record the gas volume and the water bath temperature at regular temperature intervals, for example every 5 to 10 °C, as the bath heats from room temperature toward boiling, allowing time for the trapped gas to reach thermal equilibrium with the water before each reading.
Analysing the data: Plot volume (V) against temperature in kelvin. A straight line results whose extrapolation back to V = 0 passes very close to 0 K (−273 °C), the basis of the absolute temperature scale, confirming Charles's law (V₁/T₁ = V₂/T₂).
Controlling variables
In the Boyle's law investigation, temperature must be controlled (kept constant); in the Charles's law investigation, pressure must be controlled, the plunger stays free-moving at atmospheric pressure. In both investigations, the amount of trapped gas (n) is fixed by sealing the syringe so no gas escapes.
Boyle's law is tested using a gas syringe connected to a pressure sensor at constant temperature: a graph of P against 1/V gives a straight line through the origin. Charles's law is tested using a gas syringe in a heated water bath at constant (atmospheric) pressure: a graph of V against T (kelvin) gives a straight line that extrapolates close to 0 K. Both investigations keep the amount of gas fixed by sealing the syringe.
Pause, copy the highlighted method into your book before moving on.
Did you get this? True or false: in the Boyle's law investigation, plotting pressure against 1/volume gives a straight line, but plotting pressure against volume directly gives a curve.
Sort the steps+7 XP
Put these steps of the Charles's law gas-syringe investigation into the correct order.
Submerge the syringe barrel in a water bath fitted with a thermometer.
Record the gas volume and water bath temperature at regular intervals as the bath heats.
Trap a fixed volume of air in a gas syringe with the plunger free to move.
Heat the water bath slowly and evenly, allowing the trapped gas time to reach thermal equilibrium.
Plot volume against temperature in kelvin and check the line extrapolates close to 0 K.
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Four worked examples
Worked examples · reveal as you go
Worked Example 1, Solid reactant → gas volume at STP+5 XP on full reveal
What volume of CO₂ is produced at STP when 25.0 g of CaCO₃ decomposes? CaCO₃ → CaO + CO₂. (Ca=40.078, C=12.011, O=15.999)
Click two steps to swap them. Order the gas-stoichiometry method to solve: what volume of CO₂ is produced at STP when 25.0 g of CaCO₃ decomposes? (CaCO₃ → CaO + CO₂)
Apply the mole ratio CaCO₃ : CO₂ = 1 : 1, so n(CO₂) = 0.2498 mol.
Identify the conditions (STP → use 22.71 L/mol) and confirm the equation is balanced.
Final answer: V(CO₂) = 5.67 L at STP.
Convert mass to moles: MM(CaCO₃) = 100.09; n(CaCO₃) = 25.0 ÷ 100.09 = 0.2498 mol.
Convert moles of gas to volume: V = n × Vₘ = 0.2498 × 22.71.
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The three traps that cost marks
Common errors · the 3 traps that cost marks
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Using 22.71 L/mol for RTP conditions (or 24.8 for STP)
This is the single most tested trap in gas stoichiometry. The question will almost always specify conditions, read for "STP", "standard conditions", "0°C" (use 22.71), or "RTP", "room temperature", "25°C" (use 24.8). Using the wrong value gives an answer that is off by a factor of 24.8 ÷ 22.71 = 1.107, a 10.7% error that will cost marks even if all other steps are correct.
✓ Fix: Before any calculation, underline the conditions stated in the question. Write "STP → 22.71" or "RTP → 24.8" at the top of your working before you start.
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Forgetting to convert mass to moles before applying the mole ratio
When a solid reactant mass is given and a gas volume is asked for, students sometimes skip Step 2 (n = m ÷ MM) and go straight from mass to volume using the molar volume. This is wrong, molar volume converts moles to litres, not grams to litres. You must convert mass → moles first, then apply the ratio, then convert moles → volume.
✓ Fix: Always go mass → moles → ratio → moles of gas → volume. Never skip the mass-to-moles step, even when the question asks for a gas volume.
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Applying the gas volume ratio shortcut when reactants are not all gases
The volume ratio shortcut (volume ratio = coefficient ratio) only applies when ALL species in the comparison are gases at the same temperature and pressure. In CaCO₃ → CaO + CO₂, the CaCO₃ and CaO are solids, only CO₂ is a gas. You cannot say "1 L of CaCO₃ produces 1 L of CO₂", solids don't have volumes in this sense. The shortcut works only for reactions like H₂ + Cl₂ → HCl, where all species are gases.
✓ Fix: Use the volume ratio shortcut only when every reactant and product you're comparing is explicitly a gas in the question. If any solid or liquid is involved, use the full 4-step method.
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Quick-fire drill, then revisit your thinking
Quick-fire practice · 5 reps +2 XP per reveal
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What volume does 2.0 mol of an ideal gas occupy at 25 °C and 100 kPa? (Vₘ = 24.79 L mol⁻¹)
V = nVₘ = 2.0 × 24.79 = 49.58 L.
2
How many moles of gas are in 12.4 L at 25 °C and 100 kPa? (Vₘ = 24.79 L mol⁻¹)
n = V/Vₘ = 12.4 / 24.79 = 0.500 mol.
3
For N₂ + 3H₂ → 2NH₃, what volume of H₂ reacts with 10 L of N₂ (same temperature and pressure)?
At constant T and P, volume ratio = mole ratio. H₂ : N₂ = 3 : 1 → 10 × 3 = 30 L H₂.
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What volume of CO₂ (25 °C, 100 kPa) forms when 0.20 mol of CaCO₃ decomposes? (CaCO₃ → CaO + CO₂)
n(CO₂) = 0.20 mol → V = 0.20 × 24.79 = 4.96 L.
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Find the mass of 6.0 L of O₂ at 25 °C and 100 kPa. (Vₘ = 24.79 L mol⁻¹, M(O₂) = 32.00)
n = 6.0 / 24.79 = 0.242 mol; m = nM = 0.242 × 32.00 = 7.75 g.
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Revisit your thinking
At the start of this lesson, you thought about what extra step is needed to find the volume of gas consumed or produced in a stoichiometry problem.
The answer is: gas stoichiometry is simply the 4-step method with one extra conversion. Before Step 1 (if gas volume is given), use n = V ÷ molar volume to convert to moles. After Step 3 (if gas volume is the answer), use V = n × molar volume. The molar volume is 22.71 L/mol at STP (0°C) or 24.8 L/mol at RTP (25°C), always read the conditions in the question before choosing.
Reflect: how did your initial thinking compare to what you've learned?
A fixed amount of gas at constant temperature occupies 2.0 L at 100 kPa. It is compressed to 1.0 L. Using Boyle's law (P₁V₁ = P₂V₂), what is the new pressure?
✓
Practice questions
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Multiple choice
+2 XP per correct · +5 bonus if perfect
Pick your answer, then rate your confidence. That tells the system what to drill next.
Fill the blanks+4 XP
Complete the gas-stoichiometry working below. Reaction: Zn + 2HCl → ZnCl₂ + H₂. What mass of Zn is needed to produce 3.72 L of H₂ at RTP? (Zn = 65.38)
Step 1, Identify conditions: RTP, so molar volume = L/mol.
Step 2, Gas volume → moles: n(H₂) = V ÷ Vₘ = 3.72 ÷ = mol.
Step 3, Mole ratio Zn : H₂ = : 1, so n(Zn) = 0.150 mol.
Step 4, Moles → mass: m(Zn) = n × MM = 0.150 × 65.38 = g.
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Short answer
ApplyLower-order5 MARKS
Q1. When limestone (CaCO₃) is heated in a kiln, it decomposes: CaCO₃ → CaO + CO₂. A kiln processes 500 kg of limestone per hour. (a) Calculate the volume of CO₂ produced per hour at STP. (b) Explain why the actual volume produced would differ from your calculated value. (Ca=40.078, C=12.011, O=15.999)
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AnalyseMid-order5 MARKS
Q2. 6.54 g of zinc reacts with excess hydrochloric acid: Zn + 2HCl → ZnCl₂ + H₂. (a) Calculate the volume of H₂ produced at RTP. (b) A student accidentally uses the STP molar volume. Calculate their answer and the percentage error this introduces. (Zn=65.38)
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EvaluateHigher-order3 MARKS
Q3. Explain why the volume ratio shortcut (volume ratio = coefficient ratio) can be applied to the reaction H₂ + Cl₂ → 2HCl but cannot be applied to CaCO₃ → CaO + CO₂ when finding the volume of CO₂ from a given mass of CaCO₃.
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AnalyseMid-order4 MARKS
Q4. Consider the combustion of propane: C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(g). 15.0 L of propane gas is burned completely at RTP. (a) Use the gas volume ratio (all species are gases at the same temperature and pressure) to find the volume of CO₂ produced. (b) Confirm the same answer by converting the propane to moles (n = V ÷ 24.8), applying the mole ratio, then converting back to volume. State which method is faster and why both are valid here.
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EvaluateHigher-order4 MARKS
Q5. A chemistry student claims: "It doesn't matter whether you use 22.71 or 24.8 L/mol as the molar volume, the difference is only about 9%, so it won't affect whether you pass or fail an exam question." Critically evaluate this claim with reference to a specific calculation.
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📖 Comprehensive answers (click to reveal)
Activity 2, Data Table Answers
Row 1 (2H₂O₂ → 2H₂O + O₂, STP):
MM(H₂O₂) = 34.015; n = 17.0÷34.015 = 0.4998 molRatio H₂O₂:O₂ = 2:1; n(O₂) = 0.4998÷2 = 0.2499 molV(O₂) = 0.2499 × 22.71 = 5.67 L at STP
2. 1.14 L. n(CaCO₃) = 5.00÷100.09 = 0.04996 mol; ratio 1:1; n(CO₂) = 0.04996; V = 0.04996×22.71 = 1.135 ≈ 1.14 L at STP (NESA standard 22.71 L/mol). (The older 22.4 L/mol value would give 1.12 L, which is not the keyed answer.)
3. A, 3.60 g. n(H₂) = 4.96÷24.8 = 0.2000 mol; ratio H₂:H₂O = 2:2 = 1:1; n(H₂O) = 0.2000; MM(H₂O) = 18.015; m = 0.2000×18.015 = 3.60 g.
4. D, 6.00 L. All gases at same T/P → volume ratio = coefficient ratio. H₂:NH₃ = 3:2; V(NH₃) = 9.00×(2÷3) = 6.00 L.
5. B. V = n × molar volume. Using 24.8 instead of 22.71 gives a larger multiplier → larger (incorrect) volume. Error = (24.8−22.71)÷22.71 × 100 = 9.2% too high.
6. C. n(O₂) at STP: V = n × 22.71; n = 5.60 ÷ 22.71 = 0.2466 mol. At RTP: V = 0.2466 × 24.8 = 6.12 L. The classmate's principle is correct (same moles) but the number 5.68 L is wrong, the correct RTP equivalent is 6.12 L (using NESA 22.71 L/mol).
7. A. The volume ratio shortcut (coefficient ratio = volume ratio) applies only when all reactants and products involved are gases at the same temperature and pressure. H₂ + Cl₂ → 2HCl satisfies this, all three species are gases. The other options involve solids, solutions, or liquids where the shortcut cannot be used.
Short Answer Model Answers
Q1 (5 marks):
(a) m(CaCO₃) = 500,000 g; MM(CaCO₃) = 100.09n = 500,000 ÷ 100.09 = 4996 mol; ratio 1:1; n(CO₂) = 4996 molV(CO₂) = 4996 × 22.71 = 113,459 L ≈ 1.13 × 10⁵ L at STP
(b) The actual volume would differ because: (i) the kiln operates at high temperature, not STP, gases expand at higher temperatures, so the actual volume would be much larger than calculated; (ii) the reaction may not proceed to 100% completion (yield < 100%), giving less CO₂ than theoretically predicted; (iii) the limestone may contain impurities, reducing the effective mass of CaCO₃ available.
Q3 (3 marks): The volume ratio shortcut can be applied to H₂ + Cl₂ → 2HCl because all three species in this reaction are gases at the same temperature and pressure. By Avogadro's law, equal volumes of gases contain equal numbers of moles, so the volume ratio equals the coefficient ratio directly (1 L H₂ reacts with 1 L Cl₂ to give 2 L HCl). In contrast, CaCO₃ and CaO are solids, they do not occupy measurable gas volumes. The mole concept still applies to them, but their "volume" in the Avogadro's law sense is not relevant. To find the volume of CO₂ produced, you must first convert the mass of CaCO₃ to moles (using n = m ÷ MM), apply the mole ratio, then convert CO₂ moles to volume. The shortcut cannot be used for any reaction involving solids or liquids.
Q4 (4 marks):
(a) Volume ratio (all species gases at same T, P): C₃H₈ : CO₂ = 1 : 3, so V(CO₂) = 15.0 × 3 = 45.0 L(b) n(C₃H₈) = 15.0 ÷ 24.8 = 0.6048 mol; n(CO₂) = 3 × 0.6048 = 1.815 mol; V = 1.815 × 24.8 = 45.0 L
Both methods give 45.0 L. The volume ratio shortcut is faster because, by Avogadro's law, equal volumes of gases at the same temperature and pressure contain equal numbers of particles, so the volume ratio equals the coefficient ratio directly, no molar volume conversion needed. Both are valid here only because every reactant and product compared is a gas at the same conditions.
Q5 (4 marks): The claim is incorrect. Choosing the molar volume for the wrong conditions introduces a systematic error of about 9% in every gas volume calculation. In an exam, this means the final numerical answer is wrong, typically losing the answer mark even if working is shown. Example: n = 0.500 mol of CO₂; correct V(RTP) = 0.500 × 24.8 = 12.4 L; using the STP value instead: 0.500 × 22.71 = 11.4 L, off by 1.0 L. In a 3-mark calculation, a wrong numerical answer usually means losing at least the final mark. Furthermore, if the error propagates into a subsequent calculation (e.g., finding mass from the wrong volume), multiple steps are compromised. The student should always identify the stated conditions before choosing the molar volume.
✓
Retrieve, then finish
Check what actually stuck
01
Take the full module quiz
quiz
A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.
