Question

Liquid octane (CH3(CH2)6CH3) will react with gasous oxygen (O2) to produce gasous carbon dioxide (CO2) and gasous water (H2O).

What is the theoretical yield of water formed from the reaction of 88.0 g of octane and 166.g of oxygen gas?

Be sure your answer has the correct number of significant digits in it.

Answer 1

2CH3(CH2)6CH3((g) +25O2(g) ------------------>16CO2(g) + 18H2O(g)

no of moles of C8H18   = W/G.M.Wt

                                        = 88/114.23   = 0.77 moles

no of moles of O2         = W/G.M.Wt

                                      = 166/32   = 5.1875moles

2 mole of C8H18 react with 25 moles fo O2

0.77 moles of C8H18 react with = 25*0.77/2   = 9.625 moles of O2 is required

O2 is limitin reactant

25 moles of O2 react with C8H18 to gives 18 moles of H2O

5.1875 moles of O2 react with C8H18 to gives = 18*5.1875/25   = 3.735 moles of H2O

mass of H2O = no of moles * gram molar mass

                         = 3.735*18   = 67.23g of H2O

Theoretical yield of H2O   = 67.23g of H2O >>>answer