Question

Ammonia NH3 may react with water to form oxygen and nitrogen gas .

if 2.55g of NH3 reacts with 3.83g O2 and produces .950L of N2 at 295k and 1.00atm , which reactant is limiting ?

what is the percent yeild of the reaction ?

Ammonia, NH3, may react with oxygen to form nitrogen gas and water. 4NH,(aq) +30,(8) 2N,(s) + 6H,0(1) if 2.55 g of NH3 reacts with 3.83 g Og and produces 0.950 L of N2, at 295 K and 1.00 atm, which reactant is limiting? O O NH3(aq) O2(g) What is the percent yield of the reaction? Number

Answer 1

6H₂ou 4 NH3(aq) + 302 (g) → 2N2(g) + 6 mass of NH3 = 2.55 g molare mass of NH3 = 17.031 glmol mous of NH3 = mass = 2.55 31 = 0.15 moles molau mass mass of O2 = 3.83g for molar mass of O₂ = 16 gimol moles of 02= 3.83 - 0.24 moles 16 4 moles of NH3 reacts with 3 moles of Oz 0.15 moles of NH3, moles of O2 required = 2 x0.15 = 0.1125 molis moles of O2 available > moles of Oz nequeed . NH3 (aq) acts as limiting reagent ! For Na pv=nRT. i atm x 0.95L = nxo.0821 atm. Lx molok 295k moles of Na = 0.039 moles molar mass of No = 28 g/mol mass of No produced = 1.092 g Theoretically, 4 NH3 Peroduces a produces Nr. moles 0015 NU3 peroduces 0.075 moles mass of Na expected = 0.075 X 28=2.19

yield = actual yield Theoretical yield X 100 1.092 2.1 -x 100 Percent yield = 52%