Question

Consider the combustion of butene: $$ \mathrm{C}_{4} \mathrm{H}_{8}(g)+6 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g) $$ What volume of butene at \(188^{\circ} \mathrm{C}\) and \(2.50 \mathrm{~atm}\) can be burned with \(12.0 \mathrm{~L}\) of \(\mathrm{O}_{2}\) at 745 torr and \(25.0^{\circ} \mathrm{C}\) ?

Short answer

The volume of butene that can be burned under the given conditions is 1.48 L.

Step-by-step solution

Calculating Moles of Oxygen

Firstly, calculate the number of moles of \(\mathrm{O}_{2}\) using the ideal gas law. Given that pressure is 745 torr (which is equivalent to \(745/760 = 0.979\) atm), volume is 12.0 L, temperature is \(25.0^{\circ}\mathrm{C} = 298.15\) K (remember to always convert temperature to Kelvin in gas law calculations) and \(R = 0.08206 \mathrm{L.atm/mol.K}\), we have \(n = PV/RT = (0.979 \mathrm{atm} \times 12.0 \mathrm{L}) / (0.08206 \mathrm{L.atm/mol.K} \times 298.15 \mathrm{K}) = 0.478\) mol of \(\mathrm{O}_{2}\)

Using Stoichiometry to Find Moles of Butene

From the balanced equation, it can be seen that 6 moles of oxygen react with 1 mole of butene (\(\mathrm{C}_{4} \mathrm{H}_{8}\)). Therefore, the number of moles of butene that can react with the calculated moles of oxygen is \(0.478/6 = 0.080\) mol

Calculating Volume of Butene

Finally, calculate the volume of butene using the ideal gas law again. Given that pressure is 2.50 atm, temperature is \(188^{\circ}\mathrm{C} = 461.15\) K, and we have 0.080 mol of butene, we get \(V = nRT/P = (0.080 \mathrm{mol} \times 0.08206 \mathrm{L.atm/mol.K} \times 461.15 \mathrm{K}) / 2.50 \mathrm{atm} = 1.48 \mathrm{L}\)

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental principle in chemistry and physics that relates the pressure, volume, temperature, and number of moles of a gas. The equation is given by \[ PV = nRT \]where \( P \) is the pressure of the gas, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the ideal gas constant \( (0.08206 \text{ L.atm/mol.K}) \), and \( T \) is the temperature in Kelvin. It is crucial to remember to convert all temperatures to Kelvin when using this law.In the example given, when calculating the moles of oxygen \( (\mathrm{O}_2) \), you need to apply the Ideal Gas Law adjusting the provided units correctly:

• Convert pressure from torr to atm by dividing by 760.
• Convert temperature from Celsius to Kelvin by adding 273.15.

Once everything is correctly converted, substitute the values into the equation to find the moles of the gas. This calculated value can then be used to further explore stoichiometric relationships in the chemical reaction.

Stoichiometry

Stoichiometry is the area of chemistry that involves calculating the quantities of reactants and products in a chemical reaction. This relies heavily on balanced chemical equations that depict the precise relationships between the substances involved.In the combustion of butene example, the balanced chemical equation is:\[ \mathrm{C}_{4} \mathrm{H}_{8}(g)+6 \mathrm{O}_{2}(g) \longrightarrow 4\mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g) \]This equation informs us that:

• 1 mole of butene reacts with 6 moles of oxygen.
• 4 moles of carbon dioxide and 4 moles of water are produced.

To determine how many moles of butene will react with a given amount of oxygen, like the 0.478 moles obtained, divide by the reaction stoichiometry: \(0.478 / 6\). This means you're scaling down based on the reaction ratio provided by the balanced equation.

Combustion Reaction

A combustion reaction is a type of chemical reaction where a substance reacts rapidly with oxygen, releasing energy in the form of heat and light. Combustion is common in reactions involving hydrocarbons, such as butene, which produce carbon dioxide and water as byproducts.The combustion of butene \( (\mathrm{C}_4\mathrm{H}_8) \) can be represented with the equation:\[ \mathrm{C}_{4} \mathrm{H}_{8}(g)+6 \mathrm{O}_{2}(g) \longrightarrow 4\mathrm{CO}_{2}(g)+4 \mathrm{H}_2 \mathrm{O}(g) \]This demonstrates the complete combustion of butene, where each carbon atom in the butene becomes a part of a carbon dioxide molecule, and each hydrogen atom combines with oxygen to form water. In real-life applications like engines, efficient combustion is crucial for maximizing energy output and minimizing emissions. Understanding these reactions helps in predicting how fuels will behave under various conditions.