rmine the amount of carbon dioxide produced when 2750 grams of octane reacts with excess oxygen, we need to consider the balanced combustion equation for octane and the molar ratios between octane and carbon dioxide.
The balanced combustion equation for octane (C8H18) is:
C8H18 + 12.5 O2 -> 8 CO2 + 9 H2O
From the balanced equation, we can see that 1 mole of octane produces 8 moles of carbon dioxide.
First, we need to convert the mass of octane to moles. The molar mass of octane (C8H18) is approximately 114.22 g/mol.
Number of moles of octane = 2750 g / 114.22 g/mol ≈ 24.07 mol
Now, we can determine the amount of carbon dioxide produced using the mole ratio:
Number of moles of carbon dioxide = 8 moles CO2 * (24.07 mol octane / 1 mol octane) ≈ 192.56 mol
Finally, we can convert the moles of carbon dioxide to grams. The molar mass of carbon dioxide (CO2) is approximately 44.01 g/mol.
Mass of carbon dioxide = 192.56 mol * 44.01 g/mol ≈ 8477.53 g
Therefore, approximately 8477.53 grams (or 8.48 kilograms) of carbon dioxide would be produced when 2750 grams of octane react with excess oxygen in the combustion reaction.
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