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