Determine the mass of water formed when 12.5 L NH₃ (at 298 K and 1.50 bar) is reacted with 18.9 L of O₂ (at 323 K and 1.1 bar). 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g) A) 16.7 g H₂O B) 20.7 g H₂O C) 37.7 g H₂O D) 13.8 g H₂O E) 27.9 g H₂O

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Quizplus · Accepted Answer

The balanced chemical equation is:4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)From the ideal gas law, we can calculate the number of moles of NH3 and O2:n(NH3) = (12.5 L) * (1.50 bar) / (0.0821 L bar / mol K) * (298 K) = 0.750 moln(O2) = (18.9 L) * (1.1 bar) / (0.0821 L bar / mol K) * (323 K) = 0.875 molAccording to the stoichiometry of the balanced equation, 4 moles of NH3 react with 5 moles of O2. Therefore, NH3 is the limiting reactant in this reaction.The number of moles of H2O produced is:n(H2O) = 6 * n(NH3) = 6 * 0.750 mol = 4.50 molThe mass of H2O produced is:m(H2O) = n(H2O) * M(H2O) = 4.50 mol * 18.015 g/mol = 81.07 gTherefore, the mass of water formed is 81.07 g.

Determine the Mass of Water Formed When 12