The equation of the given reaction is
2C8H18 + 25O2 --> 16CO2 + 18H2O
The mass of octane equals 0.79g/mL×1000mL=790g0.79g/mL\times1000mL=790g0.79g/mL×1000mL=790g
n(CO2)=790g(C8H18)114.2g/mol×16mol(CO2)2mol(C8H18)=55.3moln (CO_2)=\frac{790g (C_8H_{18})}{114.2g/mol}\times\frac{16mol(CO_2)}{2mol(C_8H_{18})}=55.3moln(CO2)=114.2g/mol790g(C8H18)×2mol(C8H18)16mol(CO2)=55.3mol
V(CO2)=55.3mol×22.4L/mol=1.24⋅103LV(CO_2)=55.3mol\times22.4L/mol=1.24\cdot10^3LV(CO2)=55.3mol×22.4L/mol=1.24⋅103L
Answer: 1.24⋅103L1.24\cdot10^3L1.24⋅103L
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