A 6.00-L sample of C₂H₄(g) at 2.00 atm and 293 K is burned in 6.00 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

The balanced chemical equation is:C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(g)The initial moles of C2H4 are:moles of C2H4 = (6.00 L) x (2.00 atm) / (0.0821 L atm/mol K) x (293 K) = 0.482 molThe initial moles of O2 are:moles of O2 = (6.00 L) x (2.00 atm) / (0.0821 L atm/mol K) x (293 K) = 0.482 molAccording to the balanced chemical equation, 1 mole of C2H4 reacts with 3 moles of O2. Therefore, the limiting reactant is C2H4.The moles of CO2 produced are:moles of CO2 = 2 x moles of C2H4 = 2 x 0.482 mol = 0.964 molThe moles of H2O produced are:moles of H2O = 2 x moles of C2H4 = 2 x 0.482 mol = 0.964 molThe total moles of gases produced are:moles of gases produced = moles of CO2 + moles of H2O = 0.964 mol + 0.964 mol = 1.928 molThe final volume of all gases at 2.00 atm and 293 K is:V = (nRT)/P = (1.928 mol x 0.0821 L atm/mol K x 293 K) / 2.00 atm = 2.66 LTherefore, the correct answer is A.

A 600-L Sample of C2H4(g) at 2

A 600-L Sample of C₂H₄(g) at 2