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Calculate the number of grams of carbon dioxide produced from complete combustion of one liter of octane by placing the conversions in the correct order. Make sure that the units of adjacent conversion factors cancel out.

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Answer:

2 × 10³ g

Step-by-step explanation:

Let's consider the complete combustion of octane.

C₈H₁₈(l) + 12.5 O₂(g) ⇄ 8 CO₂(g) + 9 H₂O(l)

We can establish the following relations:

  • 1 L = 10³ mL.
  • 1 mL of octane has a mass of 0.703 g (ρ = 0.703 g/mL).
  • The molar mass of octane is 114.23 g/mol.
  • The molar ratio of C₈H₁₈ to CO₂ is 1:8.
  • The molar mass of CO₂ is 44.01 g/mol.

The mass of CO₂ produced from the complete combustion of 1 L of octane is:


1LC_(8)H_(18).(10^(3)mLC_(8)H_(18))/(1LC_(8)H_(18)) .(0.703gC_(8)H_(18))/(1mLC_(8)H_(18)) .(1molC_(8)H_(18))/(114.23gC_(8)H_(18)) .(8molCO_(2))/(1molC_(8)H_(18)) .(44.01gCO_(2))/(1molCO_(2)) =2* 10^(3) gCO_(2)

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