What is the answer?

Question

Gasoline (octane) combusts with oxygen gas according to the following reaciton:

2𝐶8𝐻18+25𝑂2⟶16𝐶𝑂2+18𝐻2𝑂

Calculate the volume, in Liters of CO2 (g)  produced when 649.09 g of C8H18 react at 150.89  and 1.46 atm.  (round to hundredths place)   

Answer 1

Starting with 649.09 g of C₈H₁₈, let's calculate the molar mass so we can find the equivalent number of moles of the octane:

(From the Periodic Table)

C: 12.011 g/mol x 8 = 96.088 g.mol

H: 1.008 g/mol x 18 = 18.144 g/mol

96.088 g.mol + 18.144 g/mol = 114.232 g/mol

And now calculating the number of moles: 649.09/114.232 = 5.6822 moles of octane (C₈H₁₈).

Using the balanced chemical reaction equation, we see that there is a molar ratio of 16 moles of CO₂ produced for every 2 moles of C₈H₁₈ reacted. So if there are 5.6822 moles of octane (C₈H₁₈) reacted, then there are 5.6822(16/2) = 45.4577 moles of CO₂ produced.

The ideal gas law says PV = nRT or, since we are looking for V (volume), then:

V = nRT/P

n = 45.4577 moles

T = 150.89°C + 273.15° = 424.04 K (must use Kelvin in this equation, not ˆC)

P = 1.46 atm

R = 0.08205746 L•atm/K/mol (look this constant up in a book or on Google)

So V = (45.4577)(0.08205746)(424.04)/1.46

V = 1083.38 L