Determine the mass of water formed when 12.5 L NH₃ (at 298 K and 1.50 atm)is reacted with 18.9 L of O₂ (at 323 K and 1.1 atm). 4 NH₃(g)+ 5 O₂(g)→ 4 NO(g)+ 6 H₂O(g) A) 17.0 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:4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)The number of moles of NH3 is:n(NH3) = (12.5 L) * (1 mol / 22.4 L) = 0.558 molThe number of moles of O2 is:n(O2) = (18.9 L) * (1 mol / 22.4 L) = 0.844 molThe limiting reactant is NH3, because it produces the least number of moles of water.The number of moles of water produced is:n(H2O) = 6 * n(NH3) = 6 * 0.558 mol = 3.348 molThe mass of water produced is:m(H2O) = n(H2O) * M(H2O) = 3.348 mol * 18.015 g/mol = 60.3 gTherefore, the mass of water formed is 60.3 g.

Determine the Mass of Water Formed When 12