Question

5. 70.0 grams of propane (c'He) were burned in 65.0 g. of o2 to yield carbon dioxide and water. a. How many grams of the excess reactant remained? b. If 44.5 g. of CO2 were collected, what was the yield?
Chemistry

Answer 1

700g =1.59mol mole of CHmass molarmass 44.0g mol 65.0g203mol mol of O320gi mol bal enced eq* of combution of Propane is to find the limiting reagent each mole of reactant diveded by it's coefficent 1.59 for C3Hs 2.03-41 O. lowest number is O, >limiting reagent is O2 excess reagent is C3H。 a) from balenced eq* for 1.0mol of CzHg, requires 5.0mol of O2 hence 2.03O 2.03 0.406mol of C3Hs 5.0 → excess reagent remains = 1.59-0.406 = 1 . 1 84mol of C3H。 mass of C3Hzreamains-1.184molx44.0g/mol- 52.096g b) given actual yield of CO2-44.5g first we cal culate theoriti cal yield of CO2 from balenced eq' 5.0mol of O,, produces 3.0mol of CO, hence ?CO2 2.03x3.0 =1. 128mol orco, theoretial mass of CO2 1.128molx44.0g/mol -53.592g hence 44.5 53.592 ×100= ×100 percent yield- _actual yield theoretical yield percent yield 83.04