Chem Homework. Really need help

Question

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

2C8H18(l)+25O2(g)⟶16CO2(g)+18H2O(l)

If 611 mol of octane combusts, what volume of carbon dioxide is produced at 24.0 ∘C and 0.995 atm?

Answer 1

Hi Brianna!

Whenever a reactant is given to you and the question wants an amount of a product (whether that's grams, moles, or mL), then stoichiometry is involved. For all stoichiometry problems, you must ensure that you have a balanced chemical equation, which it appears that you do. From there, you must compare what is given versus what you can find. Since you are given moles of a reactant (octane), you can find moles of the product (CO2). Then, since the question wants a volume and you are given one pressure and one temperature, this indicates you will need ideal gas law.

Summary:

When are you are given moles of a reactant ----> you can find moles (or grams) of a product using stoich.

When given one pressure, one temperature, one volume, and/or one mole ----> use ideal gas law.

First, we always start with what is given in the question ---> 611 mol octane

611 mol C8H18 | 16 mol CO2

----------------------------------------------------------------- (To cancel moles of octane, I must put moles octane

| 2 mol C8H18 in the denominator. I can convert TO moles of CO2

by using the mole to mole ratio from the coefficients

of the balanced chemical equation)

We can multiply 611 * 16 and divide by the 2. We have 4888 mol CO2.

Now that we have moles of CO2, we can find the volume by using ideal gas law (PV=nRT).

**Remember that when dealing with gases, you need to take the temperature in Celsius and convert to Kelvin by adding 273.**

P = 0.995 atm

V = x

n = 4888 mol CO2

T = 24.0C + 273 K = 297 K

R = 0.0821 (L*atm)/(mol*K)

PV = nRT To solve for V, we can divide both sides by P

------------

P

V = (nRT)/P

V = ((4888 mol)(0.0821 L*atm/mol*K)(297K))/(0.995 atm)

119,786 L

Since all of the givens have only three sig figs, it will help to convert this to scientific notation and then round to three sig figs.

1.19786 * 10⁵ L ----> 1.20 *10⁵ L

That is the final answer. 1.20 *10⁵ L

Answer 2

First make sure the equation is balanced.

2C₈H₁₈(l) + 25O₂(g) ⟶ 16CO₂(g) + 8H₂O(l) has the same number of elements on each side so it is balanced.

The equation tells you that for every 2 moles of octane (C₈H₁₈) there are 16 moles of carbon dioxide (CO₂) produced. So if there are 611 moles of octane then there will be 611•16/2 = 4888 moles of carbon dioxide (CO₂) produced.

The ideal gas law states that PV = nRT, Solving for V we get:

V = (nRT)/P but this equation requires the use of temperature in Kelvin.

24°C = 24 + 273.15 = 297.15 K

The value of the ideal gas constant "R" is 0.08205746 L atm/K/mol

V = (4888)(0.08205746)(297.15)/0.995 = 119785 Liters but since things are given in 3 sig figs, we round to 3 sig figs:

V = 1.20 x 10⁵ Liters