hi can you help me with this question The first image is ny actual problem and the other inages is a previous answerers

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Be sure to answer all parts. Aluminum nitrite and ammonium chloride react to form aluminum chloride, nitrogen, and water. What mass of each substance is present after 96.2 g of aluminum nitrite and 70.7 g of ammonium chloride react completely? 96.2 luminum nitrite 70.7 1Xmmonium chloride 58.8 luminum chloride 37.0 hitrogen 47.6 rater < Prev 2 of 15 BE Next >

91 Reactie : Aluminkin ribili with Ammonium chloride Al (NO), +3N HULI ALCI, +3N+61,0 Moles of Al ( NO ), mass molee mass = 96:24 164-947g fast = 0.583 mil 53:49 g/mol 3 moles of whyll - 70.14 ): 322 Now, Imol of Al (No need smolung Nhull 10, 058Smolf H (NG), nud sXDs85 mol 51-749elf te maten of product deboulderly on what Clearly Al Maret Limitinguts 03 melus Nel seduces Imole nella So, 1:326 mol of which produces x 1.322 20.md CI masif Alce, 44mex 133 34 glueel 5:37 Ads @ 3wel Nhaul feme 3mele So, 1.322 kmolf 4.blºgo.me 1-322 metu na Mast of Ny 1.322 mely ab oiginal 3 meluy nhill Lams 6 meles Ho Imole of which founds by mol = 2moles of me $9, 1.322 mol An Hy el forms 1. 322 2 mol of molesale most N mass of H₂O 2.66mal x 18.02glmal moleng KoH mars mean mass 7619_ 1:356 mol

3 رام Now, Imol of Al (Nools need Smoles of Newa 10, 0.583mot fAP[NO), nud s¥o-583 met = 1749 mal will Clessly News fimiling, sactante formation of product depend only on Nhull- 3 moles of Nhull produces Imole of Alle So, 1:32 mol of Nhull produces x 1.322 ml . 0-4 med me masif Aldis melex melal mass oyulmeky 133 34g/mol 58.89 (Ads 3 why NHL m 3meen So, 1.322 knot of NHO folms 1322 meleg Mans of Ny 1:322 mely 28.0 28. olglmal 3 melul nu bams 6 moles! Ho Imole of the founds by mol = 2moles of H. o. 30, 1.372 mol NH4Cl forms ! 322 x 2 mils of H2O of mass of H₂O 2.64mol x 18.02g/mol E47:65 H20 9.2 moles of kon mass 769 56. Pagined Now, molerity of molerity of konsolation = 3:00M Melerity. molus X loco volume of silinm) so, volume firmL). = 0 melal mass 21:356 mel . moles x 1000 Modlarity 1.356 mol X1000 30 M 45 ml So, volume of solution in mL = 452mL