Question

background info: So I have the balanced equation = 2C8H18(g)+25O2(g)→16CO2(g)+18H2O(g)

and oxygen is the limiting reactant of 0.180 mol of octane is allowed to react with 0.880 mol of oxygen.

Part C How many moles of water are produced in this reaction? Express your answer with the appropriate units. View Available Hint(s) : HÅR O 2 ? H2O produced = H,0 produced = Value Units Submit Part D After the reaction, how much octane is left? Express your answer with the appropriate units. ► View Available Hint(s) CH HÅR O O ? moles of C8H18 remaining = Value Units Submit

Answer 1

268 Hip () + 250240) — 1600, 689 + 18 H20(0) Clearly 25 mol O₂ rearts with 2 mol Cg Hy to produce 18 molto 50, 0.88 mol O2 (2x0.88 kw) Cgt sy to " (18 x 288mol H2O = 0.07 ml ea = 0.634 mol Water produced = 0.634 mol to us on two conto - Octane left = 0.18-0.070 -0.110 mol