Question

The complete combustion of octane, \(\mathrm{C}_{8} \mathrm{H}_{18},\) a component of gasoline, proceeds as follows: $$ 2 \mathrm{C}_{8} \mathrm{H}_{18}(I)+25 \mathrm{O}_{2}(g) \longrightarrow 16 \mathrm{CO}_{2}(g)+18 \mathrm{H}_{2} \mathrm{O}(g) $$ (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?

Short answer

(a) 18.75 mol of O₂ are needed to burn 1.50 mol of C₈H₁₈. (b) 35.0 g of O₂ are needed to burn 10.0 g of C₈H₁₈. (c) 138240.0 g of O₂ are required to burn 15.0 gal of C₈H₁₈. (d) 121665.5 g of CO₂ are produced when 15.0 gal of C₈H₁₈ are combusted.

Step-by-step solution

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

Use the balanced chemical equation to determine the stoichiometric ratio between O₂ and C₈H₁₈. From the given equation, 2 moles of C₈H₁₈ require 25 moles of O₂. Therefore, to find the required moles of O₂ for 1.50 mol of C₈H₁₈, just use the proportion for the corresponding stoichiometric coefficients in the balanced equation. \[ \text{Moles of O}_{2} = \frac{25}{2} \times 1.50 = 18.75 \: \text{mol} \] **Step 2: Calculate mass of oxygen needed for the combustion of a given mass of octane**

(b) Mass of O₂ needed for 10.0 g of C₈H₁₈

First, find the molar mass of octane (C₈H₁₈) and oxygen (O₂): - Molar mass of C₈H₁₈: (12.01 × 8) + (1.01 × 18) = 114.23 g/mol - Molar mass of O₂: (16.00 × 2) = 32.00 g/mol Next, convert the mass of octane (10.0 g) to moles using the molar mass calculated above: \[ \text{moles of C}_{8} \text{H}_{18} = \frac{10.0 \: \text{g}}{114.23 \: \text{g/mol}} = 0.0875 \: \text{mol} \] Now, apply the stoichiometric ratio from the balanced chemical equation to find the moles of O₂ required for combustion: \[ \text{moles of O}_{2} = \frac{25}{2}\times0.0875 = 1.09375 \: \text{mol} \] Lastly, convert the moles of O₂ back to grams using its molar mass calculated earlier: \[ \text{mass of O}_{2} = 1.09375 \: \text{mol} \times 32.00 \: \text{g/mol} = 35.0 \: \text{g} \] **Step 3: Calculate mass of oxygen needed for the combustion of a given volume of octane**

(c) Mass of O₂ needed for 15.0 gal of C₈H₁₈

First, convert the volume (15.0 gal) to mass, using the density of octane provided, keeping in mind the necessary unit conversion (1 gal = 3.78541 L, 1000 mL = 1 L): \[ \text{mass of C}_{8} \text{H}_{18} = 15.0 \: \text{gal} \times \frac{3.78541 \: \text{L}}{\text{gal}} \times \frac{1000 \: \text{mL}}{\text{L}} \times 0.692 \: \frac{\text{g}}{\text{mL}} = 39497.7 \: \text{g} \] Repeat the calculations as in Step 2, using this mass to find the mass of O₂ required: \[ \text{moles of C}_{8} \text{H}_{18} = \frac{39497.7 \: \text{g}}{114.23 \: \text{g/mol}} = 345.6 \: \text{mol} \] \[ \text{moles of O}_{2} = \frac{25}{2}\times345.6 = 4320.0 \: \text{mol} \] \[ \text{mass of O}_{2} = 4320.0 \: \text{mol} \times 32.00 \: \text{g/mol} = 138240.0 \: \text{g} \] **Step 4: Calculate mass of carbon dioxide produced for the given volume of octane**

(d) Mass of CO₂ produced for 15.0 gal of C₈H₁₈

Using the moles of C₈H₁₈ calculated in Step 3, apply the stoichiometric ratio from the balanced equation to find the moles of CO₂ produced: \[ \text{moles of CO}_{2} = 8\times345.6 = 2764.8 \: \text{mol} \] Finally, convert the moles of CO₂ back to grams using the molar mass of CO₂ (12.01 + 16.00 × 2 = 44.01 g/mol): \[ \text{mass of CO}_{2} = 2764.8 \: \text{mol}\times 44.01 \: \text{g/mol} = 121665.5 \: \text{g} \]

Key concepts

Stoichiometry

Stoichiometry is like a recipe for chemical reactions. It helps us know exactly how much of each ingredient we need to make a product. In chemistry, these ingredients are reactants and products. For example, in the complete combustion of octane, a certain amount of oxygen is needed to completely burn the octane and produce carbon dioxide and water. This method is based on the balanced chemical equation, which shows the exact proportions in which chemicals react and are produced.
To solve stoichiometry problems, we use the coefficients from a balanced chemical equation. These coefficients get us the molar ratios. For octane combustion, the balanced equation is:\[2\, \text{C}_8\text{H}_{18}(l) + 25\, \text{O}_2(g) \rightarrow 16\, \text{CO}_2(g) + 18\, \text{H}_2\text{O}(g)\]It's possible to calculate how much oxygen is needed for a specific amount of octane and how much of the products are created. To do this, it's vital to start with moles because stoichiometry works at the molecular level, and moles describe amounts on this scale.

• Understand the equation: know the reactants and products.
• Identify the molar ratios: use the coefficients.
• Relate different quantities using the molar ratios.

Chemical Equation Balancing

Balancing chemical equations ensures that we obey the law of conservation of mass, meaning the number of each type of atom must be the same on both sides of an equation. This step is crucial before performing any stoichiometric calculations. Balancing requires adjusting the coefficients before each chemical formula to ensure that the number of atoms for each element is equal in both reactants and products.
In our exercise, the combustion of octane is represented by: \[2\, \text{C}_8\text{H}_{18} + 25\, \text{O}_2 \rightarrow 16\, \text{CO}_2 + 18\, \text{H}_2\text{O}\]Here, the balanced equation suggests that two molecules of octane react with 25 molecules of oxygen, producing sixteen carbon dioxide molecules and eighteen water molecules. Balancing can be tricky, but it's often helpful to:

• Start with complex molecules: balance elements that appear in complex molecules first.
• Balance one element at a time: usually start with metals, then non-metals, and save H and O for last.
• Adjust coefficients: these change to balance the equation, not the subscripts in formulas.

Molar Mass Calculation

Calculating molar mass is an essential skill in chemistry, allowing us to convert between grams and moles effectively. The molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). For octane, this involves the atomic masses of carbon and hydrogen. Using the periodic table, we add the mass of each atom in the chemical formula to find the molar mass.
For example, octane (\(\text{C}_8\text{H}_{18}\)) has:

• 8 Carbon atoms, each with an atomic mass of 12.01 g/mol
• 18 Hydrogen atoms, each with an atomic mass of 1.01 g/mol

The molar mass of octane is:\[(8 \times 12.01) + (18 \times 1.01) = 114.23 \text{ g/mol}\]Similarly, for oxygen (\(\text{O}_2\)), which has a molar mass of:\[2 \times 16.00 = 32.00 \text{ g/mol}\]These calculations are vital for converting between mass and moles, as seen in various steps of the problem.

Density and Volume Conversion

Density and volume conversions allow us to connect the physical volume of a substance with its mass. This is especially useful when working with liquids, like octane, where density is needed to find how much substance we have based on its volume. In this context, density (\(\text{g/mL}\)) describes how much mass one milliliter of the substance has.
For octane, with a density of 0.692 g/mL, the process involves:

• Converting gallons to liters: knowing that 1 gallon equals 3.78541 liters.
• Converting liters to milliliters: 1 liter is 1000 milliliters.
• Multiplying the volume by the density to find the mass: this step helps in stoichiometry to find how many moles are present.

For example, converting 15 gallons of octane to grams begins with changing gallons to milliliters, then multiplying by the density to find the mass:\[15 \text{ gal} \times 3.78541 \text{ L/gal} \times 1000 \text{ mL/L} \times 0.692 \text{ g/mL} = 39497.7 \text{ g}\]Mastering these conversions ensures accurate results when predicting product amounts or reactant needs.