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All 75 Exam Topics Step-by-Step Lesson
Practice 15-Q Timed Drill
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ARCH-C-03: Hydrocarbon Combustion Gas Volumes (Avogadro's Law), Excess O2 & Limiting Reactants

Full 15-Q Timed Drill
Step 1 of 6 • The Core Idea

ARCH-C-03: Why the 150s Classroom Method Fails Under the 90s Clock

Standard Classroom Method (Too Slow)~150s (Blows 90s Clock)
Write the balanced equation: C3H8(g)+5O2(g)→3CO2(g)+4H2O(l)\text{C}_3\text{H}_8\text{(g)} + 5\text{O}_2\text{(g)} \rightarrow 3\text{CO}_2\text{(g)} + 4\text{H}_2\text{O(l)}. By Avogadro's Law, 20 cm320\text{ cm}^3 of C3H8\text{C}_3\text{H}_8 reacts with 5×20=100 cm35 \times 20 = 100\text{ cm}^3 of O2\text{O}_2 to form 3×20=60 cm33 \times 20 = 60\text{ cm}^3 of CO2\text{CO}_2. The unreacted excess O2\text{O}_2 is 150−100=50 cm3150 - 100 = 50\text{ cm}^3. At rtp, H2O\text{H}_2\text{O} is liquid, so total gas =60+50=110 cm3= 60 + 50 = 110\text{ cm}^3 (~125s).
The Fast ShortcutTarget: 40s (Saves 110s)
**Line 1 (O2\text{O}_2 & CO2\text{CO}_2 Ratios):** 1 C3H8:5 O2:3 CO2  ⟹  O2 left=150−100=50 cm31\text{ C}_3\text{H}_8 : 5\text{ O}_2 : 3\text{ CO}_2 \implies \text{O}_2\text{ left} = 150 - 100 = 50\text{ cm}^3; CO2=60 cm3\text{CO}_2 = 60\text{ cm}^3.
Line 2 (Sum Gases at rtp): 60+50=110 cm360 + 50 = 110\text{ cm}^3 (H2O(l)=0\text{H}_2\text{O(l)} = 0) (25s).
Where Students Lose Time: In 8 past UAT-UK questions (3.56% of Chemistry), candidates lose 110s by using atomic mass 16 for o₂ gas instead of molecular mass 32. Instead, apply C4 with Maths 1 skill M3.5.
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