When acetylene, \(\mathrm{C}_{2} \mathrm{H}_{2}\), is burned in oxygen, carbon dioxide and steam are formed. A sample of acetylene with a volume of \(7.50 \mathrm{~L}\) and a pressure of \(1.00\) atm is burned in excess oxygen at \(225^{\circ} \mathrm{C}\). The products are transferred without loss to a 10.0-L flask at the same temperature. (a) Write a balanced equation for the reaction. (b) What is the total pressure of the products in the \(10.0\) - \(\mathrm{L}\) flask? (c) What is the partial pressure of each of the products in the flask?

The balanced equation for the reaction of acetylene with oxygen is given by: \(\mathrm{C}_{2} \mathrm{H}_{2} + \frac{5}{2} \mathrm{O}_{2} \rightarrow 2 \mathrm{CO}_{2} + \mathrm{H}_{2} \mathrm{O}\) Next, we'll use the Ideal Gas Law to determine the number of moles of acetylene in the given conditions (7.50 L, 1.00 atm, and \(225^{\circ} \mathrm{C}\)).

The Ideal Gas Law equation is given by \(PV = nRT\), where: - P is the pressure (1.00 atm) - V is the volume (7.50 L) - n is the number of moles - R is the gas constant (0.0821 L⋅atm/mol⋅K) - T is the temperature in Kelvin (225+273.15=498.15 K) Rearrange the equation to solve for n: \(n = \frac{PV}{RT}\) Plug in the given values and solve for n: \(n = \frac{(1.00 \ \mathrm{atm})(7.50 \ \mathrm{L})}{(0.0821 \ \mathrm{L \cdot atm/mol \cdot K})(498.15 \ \mathrm{K})} = 0.182 \ \mathrm{moles}\) Now, we'll use stoichiometry to find the moles of CO2 and H2O produced from the reaction.

From the balanced equation, the stoichiometric coefficients are: - Acetylene: 1 - Oxygen: 2.5 - Carbon dioxide: 2 - Water: 1 From the number of acetylene moles we just calculated, we can now determine the moles of carbon dioxide and water: - Moles of CO2 produced: \(0.182 \cdot 2 = 0.364 \ \mathrm{moles}\) - Moles of H2O produced: \(0.182 \cdot 1 = 0.182 \ \mathrm{moles}\) Now, we'll determine the total pressure and the partial pressures of the products in the \(10.0\)-L flask.

The Ideal Gas Law equation is still given by \(PV = nRT\). We can use this equation to determine the total pressure of the products in the \(10.0\)-L flask. To do this, we'll find the total number of moles of the products (CO2 and H2O), the given volume (10.0 L), and the temperature (498.15 K). - Total moles of products: \(0.364 + 0.182 = 0.546 \ \mathrm{moles}\) - Volume: \(10.0 \ \mathrm{L}\) - Temperature: \(498.15 \ \mathrm{K}\) Now substitute these values back into the Ideal Gas Law equation and solve for the total pressure of the products: \(P = \frac{nRT}{V} = \frac{(0.546 \ \mathrm{moles})(0.0821 \ \mathrm{L \cdot atm/mol \cdot K})(498.15 \ \mathrm{K})}{10.0 \ \mathrm{L}}= 2.23 \ \mathrm{atm}\) Finally, let's determine the partial pressures of CO2 and H2O in the flask.

Using Dalton's Law of Partial Pressures, the partial pressures of CO2 and H2O can be calculated as: - Partial pressure of CO2: \(P_{CO2} = \frac{\mathrm{moles \ of \ CO2}}{\mathrm{total \ moles \ of \ products}} \times \mathrm{total \ pressure} = \frac{0.364}{0.546} \times 2.23 \ \mathrm{atm} = 1.48 \ \mathrm{atm}\) - Partial pressure of H2O: \(P_{H2O} = \frac{\mathrm{moles \ of \ H2O}}{\mathrm{total \ moles \ of \ products}} \times \mathrm{total \ pressure} = \frac{0.182}{0.546} \times 2.23 \ \mathrm{atm} = 0.74 \ \mathrm{atm}\) In conclusion, (a) The balanced equation for the reaction is \(\mathrm{C}_{2} \mathrm{H}_{2} + \frac{5}{2} \mathrm{O}_{2} \rightarrow 2 \mathrm{CO}_{2} + \mathrm{H}_{2} \mathrm{O}\). (b) The total pressure of the products in the \(10.0\) - L flask is \(2.23\) atm. (c) The partial pressure of CO2 is \(1.48\) atm, and the partial pressure of H2O is \(0.74\) atm.