The complete combustion of octane, \(\mathrm{C}_{8} \mathrm{H}_{18}\), the main component of gasoline, proceeds as follows: \(2 \mathrm{C}_{8} \mathrm{H}_{18}(l)+25 \mathrm{O}_{2}(g) \longrightarrow 16 \mathrm{CO}_{2}(g)+18 \mathrm{H}_{2} \mathrm{O}(g)\) (a) How many moles of \(\mathrm{O}_{2}\) are needed to burn \(1.50 \mathrm{~mol}\) of \(\mathrm{C}_{8} \mathrm{H}_{18} ?\) (b) How many grams of \(\mathrm{O}_{2}\) are needed to burn \(10.0 \mathrm{~g}\) of \(\mathrm{C}_{8} \mathrm{H}_{18} ?\) (c) Octane has a density of \(0.692 \mathrm{~g} / \mathrm{mL}\) at \(20^{\circ} \mathrm{C}\). How many grams of \(\mathrm{O}_{2}\) are required to burn \(15.0 \mathrm{gal}\) of \(\mathrm{C}_{8} \mathrm{H}_{18}\) (the capacity of an average fuel tank)? (d) How many grams of \(\mathrm{CO}_{2}\) are produced when 15.0 gal of \(\mathrm{C}_{8} \mathrm{H}_{18}\) are combusted?

Step by step solution

01

(a) Moles of O₂ needed for 1.50 mol of C₈H₁₈)

For this part, we will use stoichiometry to relate moles of C₈H₁₈ and O₂ from the balanced combustion equation. From the balanced equation, $$2\ \mathrm{C}_{8}\mathrm{H}_{18}(l)+25\ \mathrm{O}_{2}(g)\longrightarrow16\ \mathrm{CO}_{2}(g)+18\ \mathrm{H}_{2}\mathrm{O}(g)$$ To find the moles of O₂ needed to burn 1.50 moles of C₈H₁₈, set up a stoichiometric ratio between moles of O₂ and moles of C₈H₁₈: $$\frac{25\ \text{moles of O₂}}{2\ \text{moles of C₈H₁₈}}=\frac{x\ \text{moles of O₂}}{1.50\ \text{moles of C₈H₁₈}}$$ Next, solve for x. (b) Grams of O₂ needed to burn 10.0 g of C₈H₁₈

02

Find moles of C₈H₁₈ from given mass

We can use the molar mass of C₈H₁₈, which is \(8(12.01 \ \mathrm{g/mol}) + 18(1.008 \ \mathrm{g/mol}) = 114.23 \ \mathrm{g/mol}\). Now convert 10.0 g of C₈H₁₈ to moles: $$\text{moles of C₈H₁₈}=\frac{10.0\ \mathrm{g}}{114.23\ \mathrm{g/mol}}$$

03

Find moles of O₂ needed using stoichiometry

Use the balanced equation and the ratio from part (a) to find the moles of O₂ needed for the moles of C₈H₁₈ found in step 1.

04

Convert moles of O₂ to grams

We can use the molar mass of O₂, which is \(2(16.00 \ \mathrm{g/mol}) = 32.00 \ \mathrm{g/mol}\). Now convert the moles of O₂ from step 2 to grams of O₂: $$\text{grams of O₂}=\text{moles of O₂}\times32.00\ \mathrm{g/mol}$$ (c) Grams of O₂ required to burn 15.0 gal of C₈H₁₈

05

Convert gallons to grams of C₈H₁₈

We can use the density of octane and convert 15.0 gal to grams: $$15.0\ \mathrm{gal} \times \frac{3.785 \ \mathrm{L}}{1 \ \mathrm{gal}}\times \frac{1000 \ \mathrm{mL}}{1 \ \mathrm{L}} \times \frac{0.692 \ \mathrm{g}}{1 \ \mathrm{mL}}$$

06

Find moles of C₈H₁₈ from grams

Use the molar mass of C₈H₁₈ found in part (b) to convert the grams of C₈H₁₈ to moles.

07

Find moles of O₂ needed using stoichiometry

Use the balanced equation and the ratio from part (a) to find the moles of O₂ needed for the moles of C₈H₁₈ found in step 2.

08

Convert moles of O₂ to grams

Use the molar mass of O₂ found in part (b) to convert the moles of O₂ from step 3 to grams of O₂. (d) Grams of CO₂ produced when 15.0 gal of C₈H₁₈ are combusted

09

Use grams or moles of C₈H₁₈ calculated in part (c)

Since we have already found the grams and moles of C₈H₁₈ in part (c), we can use those for this part.

10

Find moles of CO₂ produced using stoichiometry

Use the balanced equation to set up a ratio between moles of CO₂ and moles of C₈H₁₈: $$\frac{16\ \text{moles of CO₂}}{2\ \text{moles of C₈H₁₈}}=\frac{x\ \text{moles of CO₂}}{\text{moles of C₈H₁₈}}$$ Next, solve for x.

11

Convert moles of CO₂ to grams

We can use the molar mass of CO₂, which is \(1(12.01 \ \mathrm{g/mol}) + 2(16.00 \ \mathrm{g/mol}) = 44.01 \ \mathrm{g/mol}\). Now convert the moles of CO₂ from step 2 to grams of CO₂: $$\text{grams of CO₂}=\text{moles of CO₂}\times44.01\ \mathrm{g/mol}$$

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