Stoichiometry probelm? Any braniacs at chemistry?

Question

My car has a 13.7 gallon tank. Assuming that gasoline is entirely made from octane(C8H18), how many POUNDS of CO2 would be froduced from my car each time I use a tank of gas? How many pounds of CO2 will produce from my car each year(I drive 21,500 miles each year)?
Some equalities you may find useful
- 1 lbs- 4536 kg
1 kg-2.205 lbs.
28.35g-3.785 liters
l inch-2.54 cm
density of octane-.855g/ml
my car averages 29.7 miles/gallon

use the following equation
C8H18 + 12 1/2 O2 --> 8 CO2 + 9H2O
if u can find the answer and show work, it would be greatly appreciated. Please hlp. thx It would be nice if u can show step by step using factor labeling or something. Please show work, and if steve sees this he didnt answer both parts

Answer 1

First, convert the number of gallons in your tank to something metric--liters or milliliters would probably be best. You can use a conversion calculator online, or you can do the math; I'll use latter for the sake of showing work.

1 gallon = 3.7854 liters

13.7 x 3.785 = 51.86 liters

Since the density of octane you're given is in g/mL, then you should probably determine how many mL your tank holds by multiplying your liters figure by 1000.

51,860 mL

Now determine how many grams of octane your tank holds by setting up a proportion.

51,860 mL x 0.855 g/mL = 44,340.3 g

The mL cancels out and you end up multiplying the density and the number of mL to get the number of grams of octane your tank can carry. From here, you need to find moles to get how much CO2 your car produces per tank.

44,340.3 g x mol/114.233 g = 388.157 mol

114.233 g is the molar mass of octane, so that's why that was used in the equation. Now since you have moles of octane, use your balanced equation to find moles of CO2. The equation tells you that you'll have 8 times as many moles of CO2 as you will C8H18, so multiply the moles of octane by 8.

388.157 x 8 = 3105.26 mol

That's how many moles of CO2 your car produces per tank. To find out how many grams that is, multiply the number of moles by the molar mass of CO2 (44.001 g/mol).

3105.26 mol x 44.001 g/mol = 136,634.55 g

Divide by 1000 to get your grams to kilograms.

136.63455 kg

Now, use your conversion factor between pounds and kilograms to find out how many pounds of CO2 this is.

136.63455 kg x 2.205 lbs/kg = 301.279 lbs CO2 per tank of gas

For the second part of the problem, you'll need to find out how many tanks of gas you use each year. I think you can find this by dividing the total miles you drive by the MPG of your car (which gives you total gallons used) and then dividing again by the number of gallons in your tank (to give you the total amount of tanks used).

21,500 miles x gallon/29.7 miles = 723.906 gallons
723.906 gallons x tank/13.7 gallons = 52.840 tanks

Now, multiply the number of tanks a year by the pounds of CO2 a tank to get how many pounds are produced in a year.

52.840 tanks x 301.279 lbs/tank = 15,919.58 lbs of CO2 per year

That comes out to roughly 8 tons of CO2 per year. I hope this helps, and I hope the work makes sense.

Answer 2

The first part of the question requires you to convert 13.7 gallons of octane to to moles of octane. 13.7 gallons octane ->liters ->milliliters->grams->moles using your equalities (calculate molar mass for the last conversion).

Then use mole-mole ratio from balanced equation: 1 mole octane = 8 moles CO2

Then convert moles of CO2 into pounds of CO2. Moles CO2 ->grams->kilograms->lbs. using given equalities.

2nd question: start w/21,500 miles and end with lbs CO2

21500miles ->miles per gallon->liters->milliliters->grams-> moles->mole ratio->grams->kilograms->pounds

Once you get the first one, the second is almost identical.

Answer 3

A redox reaction is made up of 2 half reactions: 1) Reduction half reaction 2) Oxidation half reaction One way i like to remember the difference between the two is "LEO the lion does GER" "LEO" stands for "lose electron oxidation" "GER" stands for "gain electron reduction" So in a nutshell, what loses electrons is getting oxidized, so it is the reducing agent and what gains electrons is reduced, so it is the oxidizing agent. I dont really know how much explaining your asking for, but i hope this helps Good luck on your exam :)