Balanced Equation for the Complete Combustion of Octane

For the complete combustion of Octane, a balanced equation is required

This balanced equation can represent the complete combustion of Octane in O2 to make carbon dioxide (CO2), and water (H2O).
2 C8H18 + 25 O2 ? 16 CO2 + 18 H2O
Calculation of Grams and Moles of CO2
We can calculate the amount of moles or grams of CO2 from the 25.4 grams C8H18. This is done using the mole to mole ratio and the molar weight of CO2. (Kumar 2020; Singh 2020). You can calculate the moles of C8H18 produced using the 44.01 g/mol molar weight of CO2. There are 16 moles produced from C8H18 because the C8H18-CO2 molar relationship is 2:16. For every 2 moles C8H18 there will be CO2. We would get 12.7 moles from 25.4 grams C8H18. (25.4 g C8H18) / (16 mol C8H18) = 12.7 MOLE CO2. For the calculation of mass, multiply the moles of CO2 obtained by the molar mass of CO2 (i.e. 12.7 molCO2 x 44.01g/mol = 562 g CO2). C8H18 can produce 562g of CO2 from 25.4 grams.

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