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Answer to Question #67380 in General Chemistry for Jordan
Question #67380
If 402 mol of octane combusts, what volume of carbon dioxide is produced at 40.0 °C and 0.995 atm?
Expert's answer
2C₈H₁₈+25O₂->16CO₂+18H₂O
V(CO₂)=402+16/2=3216
V(CO₂)=V*22.4=72038.4L
P₀V₀/T₀=PV/T;V=P₀V₀T/T₀P=101325Pa*0.0224m³*(273+40)ᵒK=0.0267
V(CO₂)=26.7L
V(CO₂)=402+16/2=3216
V(CO₂)=V*22.4=72038.4L
P₀V₀/T₀=PV/T;V=P₀V₀T/T₀P=101325Pa*0.0224m³*(273+40)ᵒK=0.0267
V(CO₂)=26.7L
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