Ammonia,NH3,may react with oxygen to form nitrogen gas and water.
4NH3(aq) + 3O2(g) => 2N2(g) + 6H2O(l)
If 2.45gNH3 reacts with 3.68gO2 and produces 0.850LN2,at 295K and 1.00atm,which reactant is limiting?
What is percent yield of reaction?
4 NH3 + 3 O2 → 2 N2 + 6 H2O (2.45 g NH3) / (17.03056 g NH3/mol) x (2 mol N2 / 4 mol NH3) = 0.072 mol N2 (3.68 g O2) / (31.99886 g O2/mol) x (2 mol N2 / 3 mol O2) = 0.0766 mol N2 NH3 produces the least N2, so it is the limiting reactant. n = PV / RT = (1.00 atm) x (0.850 L) / ((0.08205746 L atm/K mol) x (295 K)) = 0.0351mol N2 actually
(0.0351mol) / (0.072 mol) = 0.487= 48.7% yield
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