Answer to Question #182719 in Chemistry for comel

Question

The two tubes each contain N2O gas (P = 4.92 atm, V = 3 L, T = 300 K) and NH3 gas (P = 2.46 atm, V = 3L, T = 300 K). After the reaction is complete, the measured volume of nitrogen gas is under conditions where 3.2 grams of oxygen gas with a volume of 5 liters is

Answer 1

3N₂O + 2NH₃ = 4N₂ + 3H₂O

pV = nRT

n = pV/RT

R = 0.082 L⋅atm⋅K⁻¹⋅mol⁻¹

n (N₂O) = (4.92 x 3) / (0.082 x 300) = 0.6 mol

n (NH₃) = (2.46 x 3) / (0.082 x 300) = 0.3 mol

NH₃ is the limiting reagent in this reaction. Therefore:

n (N₂) = 2 x n (NH₃) = 2 x 0.3 = 0.6 mol