Question

Ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) burns in air: $$ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) $$ Balance the equation and determine the volume of air in liters at \(45.0^{\circ} \mathrm{C}\) and \(793 \mathrm{mmHg}\) required to burn \(185 \mathrm{~g}\) of ethanol. Assume that air is 21.0 percent \(\mathrm{O}_{2}\) by volume.

Short answer

About 1683.3 liters of air are needed to burn 185g of ethanol.

Step-by-step solution

Balance the Chemical Equation

To balance the combustion reaction of ethanol, compare the number of each type of atom on both sides of the equation:Original Equation:\[\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}(l) + \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g) + \mathrm{H}_{2}\mathrm{O}(l)\]Balance carbon (C) atoms: 2 carbon atoms in ethanol (\(\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}\)), need 2 \(\mathrm{CO}_{2}\):\[\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}(l) + \mathrm{O}_{2}(g) \rightarrow 2\mathrm{CO}_{2}(g) + \mathrm{H}_{2}\mathrm{O}(l)\]Balance hydrogen (H) atoms:6 hydrogen atoms in ethanol (3 moles of \(\mathrm{H}_{2}\)), need 3 moles of water:\[\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}(l) + \mathrm{O}_{2}(g) \rightarrow 2\mathrm{CO}_{2}(g) + 3\mathrm{H}_{2}\mathrm{O}(l)\]Balance oxygen (O) atoms:There are 4 oxygen atoms in \(\mathrm{CO}_{2}\) and 3 in \(\mathrm{H}_{2}\mathrm{O}\) (total 7):\[\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}(l) + \frac{7}{2}\mathrm{O}_{2}(g) \rightarrow 2\mathrm{CO}_{2}(g) + 3\mathrm{H}_{2}\mathrm{O}(l)\]To avoid fractions, multiply the entire equation by 2:\[2\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}(l) + 7\mathrm{O}_{2}(g) \rightarrow 4\mathrm{CO}_{2}(g) + 6\mathrm{H}_{2}\mathrm{O}(l)\]

Calculate Moles of Ethanol

First, find the molar mass of ethanol (\(\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}\)) by summing the atomic masses: \(2 \times \mathrm{C} (12.01) + 6 \times \mathrm{H} (1.01) + 1 \times \mathrm{O} (16.00) = 46.08 \, \mathrm{g/mol}\).Now, calculate moles of ethanol using its mass:\[\text{Moles of Ethanol} = \frac{185 \mathrm{~g}}{46.08 \, \mathrm{g/mol}} \approx 4.015 \, \mathrm{mol}\]

Calculate Moles of Oxygen Needed

Using the balanced equation, determine the moles of \(\mathrm{O}_{2}\) needed per mole of ethanol:For 2 moles of ethanol, 7 moles of oxygen are required:\[\text{Moles of } \mathrm{O}_{2} = 4.015 \, \mathrm{mol} \times \frac{7}{2} = 14.0525 \, \mathrm{mol}\]

Calculate Volume of Oxygen

Use the ideal gas law, \(PV = nRT\), to calculate the volume \(V\) of \(\mathrm{O}_{2}\):* Convert temperature to Kelvin: \(45.0^{\circ} \mathrm{C} = 318.15 \, \mathrm{K}\)* Convert pressure to atm: \(793 \, \mathrm{mmHg} = 1.043 \, \mathrm{atm}\)* Use \(R = 0.0821 \, \mathrm{L} \, \mathrm{atm} \, \mathrm{mol}^{-1} \, \mathrm{K}^{-1}\)\[V = \frac{nRT}{P} = \frac{14.0525 \, \mathrm{mol} \times 0.0821 \, \mathrm{L \, atm \, mol^{-1} \, K^{-1}} \times 318.15 \, \mathrm{K}}{1.043 \, \mathrm{atm}} \approx 353.5 \, \mathrm{L}\]

Calculate Volume of Air Needed

Since air is 21.0% \(\mathrm{O}_{2}\) by volume, the volume of air needed is:\[\text{Volume of Air} = \frac{353.5 \, \mathrm{L}}{0.21} \approx 1683.3 \, \mathrm{L}\]

Key concepts

Combustion Reaction

Combustion reactions are an essential type of chemical reaction where a substance combines with oxygen to produce energy, often in the form of heat and light. This reaction is common with organic compounds such as hydrocarbons or alcohols like ethanol. The general form of a combustion reaction is:

• Fuel + O₂ → CO₂ + H₂O

For ethanol combustion, the initial unbalanced equation is:\[\text{C}_2\text{H}_5\text{OH} + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}\] Balancing this equation involves ensuring the number of atoms for each element is the same on both sides of the equation. This includes balancing carbon, hydrogen, and oxygen. Once balanced, it shows that burning ethanol fully in oxygen results in carbon dioxide and water, showcasing a complete combustion. When balancing equations, note that the coefficients indicating moles of each compound may need to be adjusted to whole numbers if initially fractional, to provide a balanced stoichiometric equation.

Molar Mass Calculation

Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol), and is crucial for converting between grams and moles of a substance. To calculate the molar mass of ethanol (C_2H_5OH), you add together the atomic masses of all the atoms present in the molecule:

• 2 carbon atoms: \(2 \times 12.01 = 24.02 \text{g/mol}\)
• 6 hydrogen atoms: \(6 \times 1.01 = 6.06 \text{g/mol}\)
• 1 oxygen atom: \(16.00 \text{g/mol}\)

Sum these values for ethanol: \[24.02 + 6.06 + 16.00 = 46.08 \text{g/mol}\]To convert the mass of ethanol to moles, use its molar mass: \[\text{Moles of Ethanol} = \frac{\text{mass in grams}}{\text{molar mass in g/mol}}\]In our example, the calculation results in approximately 4.015 ext{ mol} of ethanol for 185 ext{ g}. This conversion is pivotal to relate quantities in grams to chemical equations, which use moles.

Ideal Gas Law

The ideal gas law is a fundamental equation in chemistry that relates the pressure, volume, temperature, and number of moles of gas using the equation: \[PV = nRT\]where

• \(P\) is pressure,
• \(V\) is volume,
• \(n\) is the number of moles,
• \(R\) is the ideal gas constant (0.0821 L atm mol^-1 K^-1), and
• \(T\) is the temperature in Kelvin.

For gas calculations involving temperature in Celsius, always convert to Kelvin by adding 273.15. Similarly, atmospheric pressure often needs conversion from mmHg to atm (1 atm = 760 mmHg). Using these conversions in the exercise, the ideal gas law helps calculate the volume of oxygen gas needed by considering the known moles of oxygen (from balanced equation), requiring temperature and pressure adjustments.

Volume of Gas Calculations

Calculating the volume of gases combines the concepts of chemical reactions and the ideal gas law. Once the moles of a gas are determined from a balanced equation, the ideal gas law is employed to find volume under specific conditions of pressure and temperature:\[V = \frac{nRT}{P}\]In this case, the balanced chemical equation shows the moles of oxygen required, and by applying current temperature and pressure in appropriate units (Kelvin and atm), the specific volume can be calculated.Because the problem states that oxygen makes up 21% of air by volume, this percentage is used to find the total air needed. Thus, divide the calculated oxygen volume by 0.21 to get the total volume of air. The exercise shows the practical application: converting theoretical chemical data into real-world requirements for air supply, following the combustion of a specific fuel amount.