Limiting Reagent & Water Volume Trick in 60 Seconds!
❓ Concept Limiting Reagent & Water Volume Trick in 60 Seconds! How do you quickly find how much water forms in a combustion reaction without writing long stoichiometry tables ? This 60-second hack covers: ✔ Limiting reagent shortcut ✔ Quick mole-ratio conversion ✔ Mass → volume trick for liquid water ✍️ Short Explanation Let’s break the entire idea into three ultra-fast steps , using the popular combustion example: C 4 H 10 + 13 2 O 2 → 4 C O 2 + 5 H 2 O \mathrm{C_4H_{10}} + \frac{13}{2} \mathrm{O_2} \rightarrow 4\mathrm{CO_2} + 5\mathrm{H_2O} This trick works for any reaction that forms liquid water . 🔹 Step 1 — Identify the Limiting Reagent FAST Instead of computing everything, compare “ moles available : moles required ”. Example: 174 kg butane → 174000 58 ≈ 3000 \frac{174000}{58} \approx 3000 320 kg O₂ → 320000 32 = 10000 \frac{320000}{32} = 10000 Reaction needs 6.5 mol O₂ per mol C₄H₁₀ . Demand for O₂ to consume all butane: 3000 × 6.5 = 19500 ...