Question

Liquid octane CH3(CH2)6CH3 reacts with gaseous oxygen gas O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O . If 43.6g of water is produced from the reaction of 36.55g of octane and 186.9g of oxygen gas, calculate the percent yield of water. Be sure your answer has the correct number of significant digits in it.

Answer 1

CH3(CH2)6CH3 + O2 ---------------------->   CO2 + H2O

balanced equation :

2C8H18 + 25O2   -----------------> 16CO2 + 18H2O

228.44 g   800 g                             704 g       324 g

36.55 g     186.9 g                             ?               ?

here limiting reagent is octane. so product H2O produced based on that

228.44 g of octan gives -----------   324 g of H2O

36.55 g of octane gives ------------- ?? H2O

mass of H2O produced = 36.55 x 324 / 228.44

                                      = 51.84 g

theoritical yield = 51.84 g

actual yield = 43.6 g

percent yield = (actual / theoretical ) x 100

                      = (43.6 / 51.8 ) x 100

                      = 84.1%