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Pavi on HD HP Tu OWtv21Online tea X Cengage BMS oytoon-Ge x ALH chp 10 med re X NSearch resuts Ge keAssignment/takeCovalentActivity.do?locator-assignment-taketakeAssignmentSessionl.ocator-assignment-take References) Review Topics Stoichiometry Use the References to access important values i seeded for this question M For the following reaction, 12.8 grams of magnesium nitride are allowed to react with 18.2 grams of water magnesium nitride(s)+ water( magnesium hydroxide(aq)+ammonia(aq) eq grams What is the maximum amount of magnesium hydroxide that can be formed? What is the FORMULA for the limiting reagent? Veited grams What amount of the excess reagent remains after the reaction is complete? Retry Entire Group No more group attempts remain Submit Answer Previous Show Hint

Answer 1

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118 Answer: Reaction is given by. Mg3 Ng (S) + 6H₂O(l) 3 99 COH)₂ (aq) + 2NH3(aq) bel ginen mass of Mg 3 N2 (6) = 12.8gon No og moles of Mg ₂ Ng(s) = ginen mass ing molay mass in g = 12.8 9 100.9 g/mol = 0,13 moles ginen mass of H₂ole) = 18,2 g. No. g moles of H₂O = ginen massing molar mass in g = 18.2 g 18 g/mol = 1.011 mol. . for limiting reagent dinide the moles with there Stoichiometric cofficienct, which has less value is the limiting reagent. for M93 N2 0.13 = 0.13 foy Mo - 1.01 20 • Mg 3 Na and the has less value to it is limiting reagent. reacting proceed as per Mg ₂ N2 us

Mg ₂ N2 and the has less value so it is limiting reagent. reacting proceed as per Mg 3N2 us 2x0.13 Now Mg 2 N2 (8) +6H₂O(l) 3Mg (OH) Laos ) + 2NH₂ (aq) Disc Puitial 0.13 mol 0.016 mol o final o mol 1.011-6 (0.13) mol 340.13 mol mol. = 0.231 mol = 0.39 mot -0.26 mda Hence, moles of Mg(OH)2 = 0.39 ono mass a Mg(OH)2 = 0.39 x 58.32 = 22.745 grams. foronula a limiting reagent is M93N2 mass Amount excess reagent remains = No. of g moles & molar 201231 x 18 = 4.158 grams

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