If 2.75 g of N2H4 reacts and produces 0.950 L of N2, at 295 K and 1.00 atm, what is the percent yield of the reaction?
N₂H₄ (aq) + O₂ (g) → N₂ (g) + 2H₂O (l)
N₂H₄ (aq) + O₂ (g) → N₂ (g) + 2H₂O (l)
Roger the Mole
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(2.75 g N2H4) / (32.04516 g N2H4/mol) x (1 mol N2 / 1 mol N2H4) = 0.085816 mol N2H4 in theory
n = PV / RT = (1.00 atm) x (0.950 L) / ((0.08205746 L atm/K mol) x (295 K)) = 0.039244 mol N2H4 actually
(0.039244 mol) / (0.085816 mol) = 0.457 = 45.7% yield N2