the reaction of 5.0g of pentane (C5H12) with a 5.0 g of oxygen gas produces 20.4 g of CO2 what is the percent yield of this reaction?

Method 1 :

The total mass of reactants = (5.0 + 5.0) g = 10.0 g
The mass of CO₂ produced must be less than 10.0 g

Hence, the reaction can never produce 20.4 g of CO₂.


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Method 2 :

Molar mass of C₅H₁₂ = (12.0×5 + 1.0×12) g/mol = 72.0 g/mol
Molar mass of O₂ = 16.0×2 g/mol = 32.0 g/mol
Molar mass of CO₂ = (12.0 + 16.0×2) g/mol = 44.0 g/mol

Initial number of moles of C₅H₁₂ = (5.0 g) / (72.0 g/mol) = 0.06944 mol
Initial number of moles of O₂ = (5.0 g) / (32.0 g/mol) = 0.1563 mol

C₅H₁₂ + 8O₂ → 5CO₂ + 6H₂O
Mole ratio C₅H₁₂ : O₂ : CO₂ = 1 : 8 : 5

If 0.1563 mol O₂ completely reacts, C₅H₁₂ needed = (0.1563 mol) × (1/8) = 0.01953 mol < 0.06944 mol
C₅H₁₂ is in excess. Hence, O₂ is the limiting reactant/reagent.

No. of moles of O₂ reacted = 0.1563 mol
Maximum no. of moles of CO₂ = (0.1563 mol) × (5/8) = 0.09769 mol
Theoretic yield of CO₂ = (0.09769 mol) × (44.0 g/mol) = 4.30 g < 20.4 g

Hence, the reaction can never produce 20.4 g of CO₂.

Total mass of reactants ( C5H12 + O2) = 5.0g + 5.0g = 10.0g
The maximum mass of products is 10.0g at 100% yield
It is impossible to produce 20.4g of CO2 from this reaction
Please check original data.

C5H12 + 8O2 --> 5CO2 + 6H2O
72g ... + 256g so O2 is the obvious limiting reactant

5 / 72 = x / 220 .... x = 220(5/72) <<< use a calculator
this is the theoretical yield of CO2, T.

% Yield = (20.4 / T) X 100 ...use a calculator to find T, then use it again to get the final answer