Calculate the number of pounds of co2 released into the atmosphere when a 15.0-gallon tank of gasoline is burned in an automobile engine. assume that gasoline is primarily octane, c8h18, and that the density of gasoline is 0.692 g·ml–1 (this assumption ignores additives). also assume complete combustion.

266.997 pounds

The balanced equation for the combustion of octane is;

2C₈H₁₈(l) + 25O₂(g) → 16CO₂(g) + 18H₂O(g)

Assuming that gasoline is purely octane:

Amount of gasoline available is 15.0 gallon

We are required to calculate the amount of CO₂ released to the atmosphere;

This can be done in several steps;

Step 1: Mass of octane in grams

Amount of octane = 15.0 gallons

But, 1 gallon = 3785.41 ml

Therefore;

15 gallon is equivalent to;

 = 15 ×3785.41 ml

= 56,781.15 ml

We are given the density of octane (0.692 g/ml) and thus we can calculate the mass of octane.

Mass = density × volume

        = 0.692 g/ml × 56,781.15 ml

        = 39,292.56 g

Step 2: Number of moles of octane

Number of moles = Mass/ Molar mass

Molar mass of octane = 114.23 g/mol

Number of moles of Octane = 39,292.56 g ÷ 114.23 g/mol

                                               = 343.978 moles

Step 3: Moles of carbon dioxide released

From the equation;

2 moles of octane completely burns in air to yield 16 moles of CO₂

Therefore;

343.978 moles of octane produces;

= (343.978 moles/2) 16

= 2751.824 moles of CO₂

Step 4: Mass of CO₂ in pounds

1 mole of CO₂ has 44.01 g

Therefore;

2751.824 moles contains;

= 2751.824 moles × 44.01 g/mol

= 121,107.77 g

But, 1 pound = 453.592 g

Therefore;

Mass of CO₂ in pounds = 121,107.77 g ÷  453.592 g

                                      = 266.997 pounds

Therefore, the mass of CO₂ released to the atmosphere is 266.997 pounds

a c d and f

just finished the assignment

може ли да зареждате повече информация, моля