J.R. S. answered 10/28/20
Ph.D. University Professor with 10+ years Tutoring Experience
(A) C3H8 + 5O2 ==> 3CO2 + 4H2O ... balanced equation
(B) The easiest way to find limiting reagent is to divide moles of reactant by coefficient in balanced eq
moles C3H8 = 14.8 g x 1 mol/44 g = 0.336 moles C3H8 (÷ 1 = 0.336)
moles O2 = 3.44 g O2 x 1 mol O2/32 g = 0.1075 moles O2 (÷ 5 = 0.0215)
So O2 is LIMITING
(C) 0.336 moles C3H8 x 3 moles CO2 / 1 mol C3H8 = 1.00 moles CO2 produced from 14.8 g C3H8
(D) 0.1075 moles O2 x 4 moles H2O / 5 moles O2 = 0.086 moles H2O produced from 3.44 g O2
(E) % yield = actual yield/theoretical yield (x100)
Theoretical yield of CO2 = 0.1075 moles O2 x 3 moles CO2 / 5 moles O2 x 44 g CO2/mol = 2.84 g CO2
% yield = 1.22 g / 2.84 g (x100) = 43.0% yield