Question

Liquid octane (CH2(CH2),CH) will react with gaseous oxygen (02) to produce gaseous carbon dioxide (CO2) and gaseous water (H,0). Suppose 94. go octane is mixed with 200. g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits. 1g x 6 ?

Answer 1

2Cq H18 + 2502 16CO2 + 18H₂O Molar Mass of C&H 18 1Ocdane) = 1149 Molar mass of O2 = 32g Mole of (& H18 - 94 = 0.825 mol Mole of O2 = 200 = 6.25 mol 22 - - 0.625 = 0.4125 2 Mole of C8H18 18 Coefficient of C&HIR Mole of O2 Coefficient of O₂ 6.25 25 = 0.25 Mole of Ca Hig . Mole of O₂ Coefficient of C& His Coefficient of O2 O2 will be limiting Reagent. Mole I of O Reacted 25 Mole of H₂O Produced 18 → Mole of H₂o produced = = 4.5 mol 0.25x 18 Molan 90 Maximum mass of H₂O could be produced = 4.50 mod H₂O = 4.5 x 18 - 45g.