A 1.00-L sample of C₂H₄(g)at 2.00 atm and 293 K is burned in 8.00 L of oxygen gas at the same temperature and pressure to form carbon dioxide and water.If the reaction goes to completion,what is the final volume of all gases at 2.00 atm and 293 K?

Question

Quizplus · Accepted Answer

From the balanced equation: C₂H₄(g) + 3O2(g) → CO2(g) + 2H2O(g) We can see that for every 1 mole of C₂H₄ that reacts, 3 moles of O2 are required. So, we need to calculate the moles of C₂H₄ present in the initial sample: n = PV/RT = (2.00 atm × 1.00 L)/(0.0821 L·atm/mol·K × 293 K) = 0.0804 mol Since 3 moles of O2 are required for every mole of C₂H₄, we need 0.0804 × 3 = 0.2412 mol of O2. Assuming that the final volume of all gases is V, we can use the ideal gas law to find V: V = nRT/P = ((0.0804 mol C₂H₄ + 0.2412 mol O2) + nCO2 + 2nH2O) × 0.0821 L·atm/mol·K × 293 K/2.00 atm The balanced equation tells us that for every mole of C₂H₄ that reacts, 1 mole of CO2 and 2 moles of H2O are produced. So, at the end of the reaction, we have: nCO2 = 0.0804 mol C₂H₄ nH2O = 2 × 0.0804 mol C₂H₄ = 0.1608 mol Substituting these values into the equation above: V = (0.0804 mol C₂H₄ + 0.2412 mol O2 + 0.0804 mol CO2 + 0.1608 mol H2O) × 0.0821 L·atm/mol·K × 293 K/2.00 atm V = 8.00 L Therefore, the final volume of all gases at 2.00 atm and 293 K is 8.00 L, which is answer choice C).

A 100-L Sample of C2H4(g)at 2

A 100-L Sample of C₂H₄(g)at 2