The complete combustion of octane, \(\mathrm{C}_{8} \mathrm{H}_{18},\) a 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 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 combustion of 1.50 mol C₈H₁₈

According to the balanced equation, 2 moles of octane need 25 moles of O₂ for complete combustion. We have 1.50 moles of octane and need to find the moles of O₂ required. To find the moles of O₂, we will set up a ratio: \(\frac{25 \,\text{moles of } O_2}{2 \,\text{moles of } C_8H_{18}}\) Since we have 1.50 moles of octane, we will multiply the ratio by the moles of C₈H₁₈: \(1.50 \, \text{moles} \times \frac{25 \,\text{moles of } O_2}{2\,\text{moles of } C_8H_{18}}\) Solve for the moles of O₂: \(1.50 \,\text{moles} \times \frac{25}{2} = 18.75 \,\text{moles }\mathrm{O}_{2}\) Thus, 18.75 moles of O₂ are needed to burn 1.50 moles of C₈H₁₈.

02

(b) Grams of O₂ needed for combustion of 10.0 g C₈H₁₈

To find the grams of O₂, we need to convert grams of octane to moles first. The molar mass of octane is \(114.23 \, g/mol\). Moles of octane = \(\frac{10.0 \, g}{114.23 \, g/mol} = 0.0875 \, mol\) Now, use the ratio from part (a) to find the moles of O₂ needed: \(0.0875 \, \text{moles} \times \frac{25 \,\text{moles of } O_2}{2\,\text{moles of } C_8H_{18}} = 1.094 \,\text{moles }\mathrm{O}_{2}\) Now, we need to convert moles of O₂ to grams. The molar mass of O₂ is \(32 \, g/mol\). Grams of O₂ = \(1.094 \, \text{moles} \times 32 \, g/mol = 35.01 \, g\) Thus, 35.01 g of O₂ are needed to burn 10.0 g of C₈H₁₈.

03

(c) Grams of O₂ needed for combustion of 15.0 gal C₈H₁₈

First, we need to convert 15.0 gal of octane to grams. We know that 1 gal = 3.78541 L and the density of octane is \(0.692 \, g/mL\). Therefore, 15.0 gal * 3.78541 L/gal * 1000 mL/L * 0.692 g/mL = 39376.87 g of C₈H₁₈ Now, convert grams of C₈H₁₈ to moles. Moles of C₈H₁₈ = \(\frac{39376.87\, g}{114.23 \, g/mol} = 344.71 \, mol\) Now, use the ratio from part (a) to find the moles of O₂ needed: \(344.71 \, \text{moles} \times \frac{25 \,\text{moles of } O_2}{2\,\text{moles of } C_8H_{18}} = 4313.38 \,\text{moles }\mathrm{O}_{2}\) Now, convert moles of O₂ to grams. Grams of O₂ = \(4313.38 \, \text{moles} \times 32 \, g/mol = 137,827.94 \, g\) Thus, 137,827.94 g of O₂ are needed to burn 15.0 gal of C₈H₁₈.

04

(d) Grams of CO₂ produced from combustion of 15.0 gal C₈H₁₈

According to the balanced equation, 2 moles of octane produce 16 moles of CO₂. We have 344.71 moles of octane from part c, so we will use this to find the moles of CO₂ produced: Moles of CO₂ = \(344.71 \, \text{moles} \times \frac{16\,\text{moles of } CO_2}{2\,\text{moles of } C_8H_{18}} = 2757.68 \,\text{moles }\mathrm{CO}_{2}\) Now, convert moles of CO₂ to grams. The molar mass of CO₂ is \(44 \, g/mol\). Grams of CO₂ = \(2757.68 \, \text{moles} \times 44 \, g/mol = 121339.92 \, g\) Thus, 121,339.92 g of CO₂ are produced when 15.0 gal of C₈H₁₈ are combusted.

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