Determine the amount of carbon dioxide that will be released when burning 121 g. glucose if the yield is 72% of theoretical.

Given:
m (C6H12O6) = 121 g
W (practical) = 72%
To find:
V (CO2) =?
Decision:
1) Let’s compose the equation of the chemical reaction
С6Н12О6 + 6О2 = 6СО2 + 6Н2О
2) We know the mass of glucose, and let CO2 be xl.
M (C6H12O6) = 12 x 6 + 12 + 16 x 6 = 180 g / mol.
Vm (CO2) = 22.4 x 6 = 134.4 L / mol.
3) 121 g: 180 g / mol = x L: 134.4 L / mol
x = (121 x 134.4): 180 = 90.3 liters – theoretical yield 100%.
4) 90.3 l = 100%
x l = 72%
x = (90.3 x 72): 100 = 65 l – practical yield 72%.
Answer: V (CO2) = 65 l.