What volume of air is needed to burn an entire 60-L (approximately 16-gal) tank of gasoline? Assume that the gasoline is pure octane, C8H18. Hint: air is 20% oxygen by volume, 1 mol of a gas occupies about 25 L at room temperature, and the density of octane is 0.70 g/cm3.

To calculate the volume of air needed to burn the entire tank of gasoline, we need to determine the stoichiometric ratio between gasoline and oxygen.

1. We know that gasoline is composed primarily of octane (C8H18). To determine the amount of oxygen required for complete combustion, we need to balance the chemical equation for burning octane:

C8H18 + 12.5O2 → 8CO2 + 9H2O

2. From the balanced equation, we can see that 1 mole of octane reacts with 12.5 moles of oxygen to produce 8 moles of carbon dioxide and 9 moles of water.

3. Now let's calculate the amount of octane in the given 60 L tank of gasoline:
- The density of octane is given as 0.70 g/cm3, which is equivalent to 700 kg/m3.
- So, the mass of the gasoline can be determined by multiplying the density with the volume:
Mass = Volume * Density = 60 L * 700 kg/m3 = 42000 kg.

4. Next, we need to convert the mass of octane to moles. The molar mass of octane (C8H18) is:
Molar mass = (12.01 g/mol * 8) + (1.008 g/mol * 18) = 114.23 g/mol.
Moles of octane = Mass / Molar mass = 42000 kg / 114.23 g/mol = 366.97 mol.

5. Based on the stoichiometric equation, for every 1 mole of octane, we need 12.5 moles of O2. Therefore, the amount of oxygen required can be calculated as:
Moles of O2 = Moles of octane * 12.5 = 366.97 mol * 12.5 = 4587.12 mol.

6. Since we know that air is approximately 20% oxygen by volume, we can calculate the volume of air needed using the ideal gas law. Considering 1 mol of a gas occupies about 25 L at room temperature and pressure (standard conditions):
Volume of air = Moles of O2 / Oxygen content in air = 4587.12 mol / 0.20 = 22935.6 mol.
Volume of air = 22935.6 mol * 25 L/mol = 573,390 L.

Therefore, approximately 573,390 L of air is needed to burn an entire 60-L tank of gasoline.

I think you can look at this problem like this.

Each year it is 2,16 * 10^13 kg more CO2 put into the atmosphere, and this is
2,16*10^13/4,5*10^15 = 0,005 increase.

So if you use a simple interest formula you should get the right answer. (same growth each year)

4,5 * 1,005^x = 9 (10^15 cancels out)
1,005^x = 2
x*log(1,005) = log(2)
x = log(2)/log(1,005)
x = 139

So it would take 139 years for the CO2 amount to double in the atmosphere, with the current use of fossil fuel.