Question

1. Given the following reaction: 4 NH3 + 50 →4 NO + 6 HOH a. What is the theoretical yield of water if you do the reaction with 13.4 g of NH, and 15.68 07? b. Assuming unlimited oxygen, how much NH, must you use to make 2. How many grams of NaNO, and how many ml of water do you need to make 2.3M NaNO3? 3. What is the net ionic reaction for the following? Pb(NO3h () KCL (9) ► 4. Given a 56 M solution of Al(NOJ, what is the concentration of oxygen atoms? Focus 3 --

Answer 1

68 Given reaction is 4NH3 + 5024 NO & GHOH 689 160g . 1089 @ NH3 = 13.49 6 O₂ = 15.6 g find limiting reagent 689 NH3 reacts with 160 goz in 13.4 g NH₃ reacts with = 160 x 13.48 531529 Hence tiniting seagent - O2 Now : 160g On produces = 108 g H₂O a. 15.6 g 0 prodace = 108 x 1506 g H₂O E 10.53 g H₂O . 108g of H₂O is produced by = 68 g of NH3 .: 65.7 g of H₂O is produced by = 68 x 65+7 g of NH₃ - 41.36 g NH3 2) 45.5m2 of 2.3 m NaNO₃ " In 1000 mL of water = 2.3485g NaNO3 present In 45.5 m2 of water - 2.3485 x 4.5 g NaNo, present 1000 - 8.89529 NaNO₃ 160 108 . g of NANDE - 8-8952 g mL of Glater = 45.5mL

☺ Net ionic reaction Pb 109.) + 2N05 (296) + ktcaq.3 + ci(29.) -> pbcla(s) + 2K109) +2 N03 (290) 6 5.6m of AI (NO3)3 No. of moles of oxygen atom = 5.6x3x3 = 50.4 mole i concentration of oxygen atom - 5004 mole 1000 m2 = 000504 mol mil or, - 50.4 mole F 50.4m