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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 O鈧� needed for combustion of 1.50 mol C鈧圚鈧佲倛

According to the balanced equation, 2 moles of octane need 25 moles of O鈧� for complete combustion. We have 1.50 moles of octane and need to find the moles of O鈧� required. To find the moles of O鈧�, 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 C鈧圚鈧佲倛: \(1.50 \, \text{moles} \times \frac{25 \,\text{moles of } O_2}{2\,\text{moles of } C_8H_{18}}\) Solve for the moles of O鈧�: \(1.50 \,\text{moles} \times \frac{25}{2} = 18.75 \,\text{moles }\mathrm{O}_{2}\) Thus, 18.75 moles of O鈧� are needed to burn 1.50 moles of C鈧圚鈧佲倛.

02

(b) Grams of O鈧� needed for combustion of 10.0 g C鈧圚鈧佲倛

To find the grams of O鈧�, 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 O鈧� 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 O鈧� to grams. The molar mass of O鈧� is \(32 \, g/mol\). Grams of O鈧� = \(1.094 \, \text{moles} \times 32 \, g/mol = 35.01 \, g\) Thus, 35.01 g of O鈧� are needed to burn 10.0 g of C鈧圚鈧佲倛.

03

(c) Grams of O鈧� needed for combustion of 15.0 gal C鈧圚鈧佲倛

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 C鈧圚鈧佲倛 Now, convert grams of C鈧圚鈧佲倛 to moles. Moles of C鈧圚鈧佲倛 = \(\frac{39376.87\, g}{114.23 \, g/mol} = 344.71 \, mol\) Now, use the ratio from part (a) to find the moles of O鈧� 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 O鈧� to grams. Grams of O鈧� = \(4313.38 \, \text{moles} \times 32 \, g/mol = 137,827.94 \, g\) Thus, 137,827.94 g of O鈧� are needed to burn 15.0 gal of C鈧圚鈧佲倛.

04

(d) Grams of CO鈧� produced from combustion of 15.0 gal C鈧圚鈧佲倛

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Combustion Reactions

The concept of combustion reactions is essential in understanding how fuels are burned to release energy. A combustion reaction involves a substance, typically a hydrocarbon, reacting with oxygen to produce carbon dioxide, water, and energy. In the case of octane,

• The balanced reaction is: \[2 \mathrm{C}_{8} \mathrm{H}_{18} + 25 \mathrm{O}_{2} \rightarrow 16 \mathrm{CO}_{2} + 18 \mathrm{H}_{2} \mathrm{O}\]
• This reaction shows that octane reacts with a large amount of oxygen to produce carbon dioxide and water, typical of hydrocarbons.

Combustion reactions are exothermic, meaning they release heat, which is why they are used for generating energy in engines. The ability to balance these reactions is crucial, as it helps us understand the stoichiometric relationships between reactants and products.

Molar Mass Calculations

Molar mass calculations are fundamental in stoichiometry, allowing us to convert between mass and moles of a substance. The molar mass of a compound is the mass in grams of one mole of its molecules.

For octane (\(\mathrm{C}_{8} \mathrm{H}_{18}\)):

• Calculate its molar mass by adding the atomic masses of all atoms in the molecule.
• Carbon: 12 g/mol, Hydrogen: 1 g/mol.
• Molar mass = \(8 \times 12 + 18 \times 1 = 114.23 \, g/mol\).

This calculation allows us to determine the number of moles contained in a given mass of octane. By converting grams into moles, we can apply stoichiometric coefficients from the balanced chemical equations to further find out how much of another substance, like oxygen, is needed or produced.

Chemical Equations

Chemical equations are symbolic representations of chemical reactions. They must be balanced to reflect the law of conservation of mass, meaning atoms are neither created nor destroyed.

To balance a chemical equation, such as the combustion of octane:

• Write down the number of atoms for each element on both sides of the equation.
• Adjust the coefficients to ensure the same number of each type of atom on both sides.

In the given reaction, the coefficients 2, 25, 16, and 18 ensure each type of atom is balanced:

• Carbon: 16 on both sides.
• Hydrogen: 36 on both sides.
• Oxygen: Balanced by adjusting the number of \(\mathrm{O}_{2}\) molecules.

Balancing equations allows us to accurately understand the proportions of reactants and products in a reaction, which is crucial for quantitative predictions in chemical processes.