Question

Hydrogen and oxygen react to form water vapor as seen in the unbalanced equation below. In a 475 ml container at 533C, hydrogen has a pressure of 0.998 atm. Inside that vessel, oxygen is contained in a small tube with a volume of 25.0 ml, same temperature, and a pressure of 2.75 atm. The small vessel of oxygen is shattered and allowed to react according to the equation H2(g) + O2(g) - HO(g) What is the limiting reactant? What is the pressure of water produced? What is the excess reactant? What is the pressure produced from the excess For credit enter the pressure of the excess reactant Submit response

Answer 1

from the Ideal gas equation PV = nRT n = PV/RT 0.998 atmx (475L S533°C = 533+2737 = 806K S 0.0821 Liatm motlu x 800k = 7.16 x 103 mol = 1.039 x 103 mot 0 2.75 atm x (25) 0.0821 Latmmor'' x 800K The Balanced equation is - 2H₂(g) + O2(g) + 2H₂O(g) of M2 : O₂ = 2:1 – required Or = 5 (7.16 x 10-3 – 2.584103 mol But we only have 1.039810-3 mol of O₂ Hence do is limiting reagent. 22 (9) & Olg) - 2H₂0 (9) 7.16 x 10 ml 1.039x109 mole + 2x 1,0398103 mol - 2X1.139x10?mo) -1039X 70 mol 2.0788 10-3 mol E (7.16 - 2.078) x 103 mol = 5.062 x 10 mol Now again by gas equation - PH₂O = H2OX RT Pro = 2.078 x103 x 0. 0821 X 806 atm (475/1000) = 0.209 atm

H₂ is in excess. PH₂ excess m 51082 x 103 x 0.0821 X 806 (475/1000) 0.706 atm] Haexcess = 6 mol fraction of H = 7.16 x 10-3 7.16 +1,0399 x 10-3 moles of H₂ ) Total moles & XH₂ = 0.873