The complete combustion of octane,a component of gasoline,is represented by the equation:
2 C₈H₁₈(l) + 25 O₂(g) → 16 CO₂(g) + 18 H₂O(l)
How many liters of CO₂(g) ,measured at 63.1 °C and 688 mmHg,are produced for every gallon of octane burned?

Question

The complete combustion of octane,a component of gasoline,is represented by the equation: 2 C<sub>8</sub>H<sub>18</sub>(l) + 25 O<sub>2</sub>(g) → 16 CO<sub>2</sub>(g) + 18 H<sub>2</sub>O(l) How many liters of CO<sub>2</sub>(g) ,measured at 63.1 °C and 688 mmHg,are produced for every gallon of octane burned? A) 7.48 L B) 7.11 × 10<sup>2</sup> L C) 5.68 × 10<sup>3</sup> L D) 5.68 L E) 1.07 × 10<sup>3</sup> L

Quizplus · Accepted Answer

The balanced chemical equation shows that for every 16 moles of CO2 produced, there are also 12 moles of CO produced. Therefore, the ratio of CO2 to CO is 16:12, which simplifies to 4:3. This means that for every 4 L of CO2 produced, there are 3 L of CO produced.

We need to find the volume of CO produced when 10 moles of octane are burned. From the balanced chemical equation, we can see that 1 mole of octane produces 8 moles of CO2 and 6 moles of CO. Therefore, 10 moles of octane will produce 80 moles of CO2 and 60 moles of CO.

Using the ratio determined earlier, we can calculate the volume of CO produced:

60 moles CO x (22.4 L/mol) = 1344 L CO

Therefore, the answer is C, 5.68 x 10^13 L CO.

The Complete Combustion of Octane,a Component of Gasoline,is Represented by the Equation