Question

RERUM ET NEM For the following reaction, 9.52 grams of ammonia are allowed to react with 15.2 grams of oxygen gas. ammonia() + oxygen() — nitrogen monoxide() + water(g) What is the maximum amount of nitrogen monoxide that can be formed? grams What is the FORMULA for the limiting reagent? grams What amount of the excess reagent remains after the reaction is complete? Submit Answer Retry Entire Group B more group attempts remaining

Answer 1

2NH₂ + soron ENO ig, +370 gs. weight of NH₃ = 9.52g. no. of mole of NH₃ = 9.529 = 0.56 mol 179lmol weight of O₂ = 15.2g. no. of mode of O₂ = 15.28 = 0.475 mol 22 2 Ace to balance egr 2 mole NH₃ react with so made of O2, then 0.56 mol wis react with 6.56x5 = 0.7 mol of O, but we have only 0.475 mol O here, I is limiting reagent. Hence, The formula for limiting reagent is O₂ as 512mose O2 produce a mole No. then, 0.475 mole O2 produce 2x0.475 = 0.38 mol of No. Merce, no. of mole of No produced = 0.38 mol. then mais of no produced = 0.38 molx 30g mod Hence, the maximum amount of nitrogen monoxide forned=114 After son complete NM remain in n so NH₃ is exces reagent. 5/2 no. of mode of excers NH₂ = 0.56 mol mol react = 0.56m09-0.475x2 mol = 0.18 mol mars of excess NH₂ = 0.18 mal x 179lmal - 3.06g. Hence, 3.obg of excers NH₃ remain after son is completed.