Could I have some help with this chemistry problem?

Question

The combustion of octane,C8H18, proceeds according to the reaction shown.

2C8H18(l)+25O2(g)⟶16CO2(g)+18H2O(l)

If 5.70×10^2 mol of octane combusts, what volume of carbon dioxide is produced at 34.0 ∘C and 0.995 atm?

Answer 1

2C₈H₁₈(l) + 25O₂(g) ==> 16CO₂(g) + 18H₂O(l) .. balanced equation

Using the stoichiometry of the balanced equation and dimensional analysis, we find moles of CO2 formed:

5.70x10² mols C₈H₁₈ x 16 mols CO₂ / 2 mols C₈H₁₈ = 4560 mols CO₂ formed

Now, to convert this to volume, we use the ideal gas law PV = nRT and solve for V (volume)

P = pressure = 0.995 atm

V = volume in liters = ?

n = moles of CO₂ = 4560 mols

R = gas constant = 0.0821 Latm/Kmol

T = temperature in K = 34 + 273 = 307K

V = nRT / P = (4560)(0.0821)(307) / 0.995

V = 115,511 L = 116,000 L (3 sig.figs.)

Answer 2

This question is a 2-step problem. Since you have a balanced equation, you will need to use mole ratios. Then you can use the Ideal Gas equation to find your volume.

1. Mole ratio step: the two substance in your equation are octane and carbon dioxide. According to your balance equation, you 2 moles of octance will make 16 moles of CO₂. This means that every mole of octane produces 8 moles of CO₂. So 5.70 x 10² molves of octance will make 9 times as much CO₂. (5.70 x 10²) x 9 = 5130 moles CO₂
2. Ideal Gas equation: PV = nRT. Since you want to find volume (V). you can divide both sides of the equation by P to get V = nRT/P, where n = 5.130 moles, R = 0.0821 atm•L/mol•K, T = 307 K (convert from °C by adding 273), and P = 0.995 atm.

V = (5130)*(0.0821)*(307) / 0.995 all of your units will cancel except the liters (L) which is volume!

Be sure to round your answer to 3 significant figures.