What is the percent yield if 100 g of C3H8 combusts to produce 200 g CO2?

Given:

mass of propane, C3H8, mp = 100 g
mass of carbon dioxide, CO2, mc = 200 g (actual yield)

Solution:

Combustion reaction Stoichiometry:

C3H8 + 5O2 -----> 3CO2 +4H2O
44.1 g/mol + 32 g/mol ---> 44.01 g/mol + 18 g/mol

Theoretical yield of carbon dioxide is calculated as follows:

Mass to mass conversion:

Theoretical yield =100 g C2H8 * (1 mol C3H8/44.1 g/mol)*(3 mol CO2 / 1 mol C3H8)*(44.01 g/mol/1 mol CO2)
Theoretical yield = 299.39 g CO2

Percent yield = (actual yield/ theoretical yield)*100 %

Percent yield = (200 g/ 299.39 g)*100 %
Percent yield = 0.6680*100 %
Percent yield = 66.8 %

C3H8 + 5O2 --> 4H2O + 3CO2
m(CO2) = 200g
n(CO2) = 200g / 44 g/mol = 4.5 mol
n(C3H8) = 4.5 mol /3 = 1.5 mol
m(C3H8) = 1.5 mol * 44 g/mol = 66.6 g (C3H8 and CO2 happen to both have a molecular mass of 44 g/mol)
Percentage yield of C3H8 is 66.6/100g = 66.6%

Equation:
C3H8 + 5 O2 = 3 CO2 + 4 H2O
1 mol C3H8 will produce 3 mol CO2
Molar mass C3H8 = 44g/mol
Mol C3H8 in 100g = 100/44 = 2.273 mol C3H8
This will produce 2.273*3 = 6.818 mol CO2
Molar mass CO2 = 44g/mol
Mass of 6.818 mol CO2 = 6.818*44 = 300g CO2 Theoretical yield
Actual yield = 200g CO2
% yield = 200/300*100 = 66.6% yield.