Question

2 C₈H₁₈ (l)    +    25 O₂ (g) ---->   16 CO₂ (g)    +     18 H₂O (g)

(OR 2 C₈H₁₈ (l)    +    25 O₂ (g) right arrow   16 CO₂ (g)    +     18 H₂O (g))

Use the following conversions: 1 mol = MW in g (2 d.p.) = 22.414 L

All gases are at STP.   How many liters of CO₂ would be produced from the reaction of 5.0 L of C₈H₁₈ with 55 L of O₂?

Theoretical yield of CO₂ from 5.0 L C8H18 = _____________

Theoretical yield of CO₂ from 55 L O2 = _____________

Limiting reagent = ____________(C₈H₁₈ or O₂ or CO₂)

Answer 1

2 C₈H₁₈ (l)    +    25 O₂ (g) ---->   16 CO₂ (g)    +     18 H₂O (g)

2 moles of C8H18 react with 25 moles of O2

2*22.414L of C8H18 react with 25 *22.414L of O2

5L of C8H18 react with = 25*22.414*5/(2*22.414)   = 62.5L of O2 is required

O2 is limiting reagent

25 moles of O2 react with excess of C8H18 to gives 16 moles of CO2

25*22.414L of O2 react with excess of C8H18 to gives 16*22.414L of CO2

55L of O2 react with excess of C8H18 to gives = 16*22.414*55/(25*22.414)   = 35.2L of CO2

Theoretical yield of CO₂ from 55 L O2 = 35.2L of CO2 >>>>answer