Stoichiometry Trick: How Much Water Forms When Butane Burns? 🔥
❓ Question Butane reacts with oxygen to produce carbon dioxide and water following the equation: C 4 H 10 ( g ) + 13 2 O 2 ( g ) ⟶ 4 C O 2 ( g ) + 5 H 2 O ( l ) If 174.0 kg of butane is mixed with 320.0 kg of O₂, the volume of liquid water formed (in liters, nearest integer) is ________. 🖼️ Question Image ✍️ Short Solution We solve by: Converting given masses to moles. Identifying the limiting reagent using reaction stoichiometry. Using stoichiometry to find moles (and mass) of H₂O produced. Converting mass of liquid water to volume (density ≈ 1.00 g·mL⁻¹). Step 1 — Molar masses & moles Molar mass (approx): M C 4 H 10 = 4 ( 12.01 ) + 10 ( 1.008 ) = 58.12 g \mol − 1 M_{\mathrm{C_4H_{10}}}=4(12.01)+10(1.008)=58.12\ \text{g·mol}^{-1} M O 2 = 32.00 g \ mol − 1 M_{\mathrm{O_2}}=32.00\ \text{g·mol}^{-1} M H 2 O = 18.015 g \ mol − 1 M_{\mathrm{H_2O}}=18.015\ \text{g·mol}^{-1} Moles of butane: n C 4 H 10 = 174000 g...