Question

The balanced equation for the combustion of butane is: 2C4H10 +13O2-->8CO2 + 10H2)

If the combustion of 42.0 g of C₄H₁₀ produces 103 g of CO₂. What is the percent yield of the reaction? (Assume oxygen is in excess.)

Answer 1

solution:

the balanced combustion reaction is

2 C₄​H₁₀​ + 13 O₂​ -----------------------> 8 CO₂ ​+ 10 H₂​O

The reaction suggests 2 moles of C₄​H₁₀ in presence of   13 moles of oxygen gives 8 moles of CO₂ and 10 moles of water.

since oxygen is excess, C₄​H₁₀ will be limiting reactant

moles = mass/Molar mass,

molar masses ( g/mole) : C₄​H₁₀  = 58g/mol, CO₂ = 44 g/mol and H₂O = 18 g/mol

therefore, Moles of C₄​H₁₀ = 42 / 58 = 0.724 moles

since 2 mole of C₄​H₁₀ gives 8 moles of CO₂

0.0.724 moles of C₄​H₁₀ will gives 0.724*(8/2) = 2.89 moles of CO₂,

So, mass of CO₂ = 2.89*44 = 127.45 gm

Mass of CO₂ produced theoretically = 127.45 gm ,

And, actual yield = 103 gm

therefore,

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

percent yield = 103/127.45 = 80.8%

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