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Answer H92L We (A) Al₂O3 + 3H₂SO4 8kg-mol Al₂O3 and mal IMAho react with Mol 1₂2504 8x3 = 24 kg-mol thSony we required only 24kg and limiting excess reactant is Hysoy reactant Abo That means above disscasion 8kg-mole react with 24 kg-mol H₂SO4 We have 40 kg- mole mole of excess percentage = of theor H₂SO4 8 kg-mol Al₂o reacting with have 40 kg-mol H₂₂ Sou that means will be excess percentage I th Sou → Al₁₂ (S0₂₁)3 + 3 tho 40kg-mol th 50₂ react in 3 mol In sou will be 8kg-mol = 8000 mol mass of AL₂O3 Mass of 8 kg md Al2(104)3= - reaction efficiency = freaction efficiemy T = while = 4014-mal excess mol of the say Total Mol of the son 40% So remain mole or excess 24 kg-mol mixture 1 *100 819-mol A/₂ 0₂ Convert completely. so we know that Abo, is limiting reactant and limiting reactant és completely convest in to product at the end of the reaction, that means 16 Ку-т-1 yoly-md 814 Conversion pescentige of Ako -XIOD 100% 818. (c) 8-kymol Aboz produce = 8 19- and Alat Sou) by storchrometry of the reaction because 1-mole Abo produce 8 ty-mol A₂₂0₂ produced mole x moleculas mass = goromol x 1029/² govo mol x 342% / mal Mass of Ah₂2 03 (Input) les Mass of Ab(Son) (auto)ing 298257 x100= = = 16kg-und नाळ Imol Al2(SO4)3 8 lg-mol Ah (Sur) } 816 ку 2736 к Al Свси) з 81619 ³x100 27365

Answer H92L We (A) Al₂O3 + 3H₂SO4 8kg-mol Al₂O3 and mal IMAho react with Mol 1₂2504 8x3 = 24 kg-mol thSony we required only 24kg and limiting excess reactant is Hysoy reactant Abo That means above disscasion 8kg-mole react with 24 kg-mol H₂SO4 We have 40 kg- mole mole of excess percentage = of theor H₂SO4 8 kg-mol Al₂o reacting with have 40 kg-mol H₂₂ Sou that means will be excess percentage I th Sou → Al₁₂ (S0₂₁)3 + 3 tho 40kg-mol th 50₂ react in 3 mol In sou will be 8kg-mol = 8000 mol mass of AL₂O3 Mass of 8 kg md Al2(104)3= - reaction efficiency = freaction efficiemy T = while = 4014-mal excess mol of the say Total Mol of the son 40% So remain mole or excess 24 kg-mol mixture 1 *100 819-mol A/₂ 0₂ Convert completely. so we know that Abo, is limiting reactant and limiting reactant és completely convest in to product at the end of the reaction, that means 16 Ку-т-1 yoly-md 814 Conversion pescentige of Ako -XIOD 100% 818. (c) 8-kymol Aboz produce = 8 19- and Alat Sou) by storchrometry of the reaction because 1-mole Abo produce 8 ty-mol A₂₂0₂ produced mole x moleculas mass = goromol x 1029/² govo mol x 342% / mal Mass of Ah₂2 03 (Input) les Mass of Ab(Son) (auto)ing 298257 x100= = = 16kg-und नाळ Imol Al2(SO4)3 8 lg-mol Ah (Sur) } 816 ку 2736 к Al Свси) з 81619 ³x100 27365

Chemistry
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Answer H922 We (A) Al₂O3 + 3H₂SO4 8kg-mol Al₂O3 and mal IMAho react with Mol 1₂2504 8x3 = 24 kg-mol throu we required only 24kg excess reactant is Hesoy and limiting reactant Abo That means above disscasion 8kg-mole react with 24 kg-mol H₂SO We have mole of - excess percentage = of theor Hasou 8 kg-mol Al₂o reacting with have 40 kg-mol H₂₂ Sou that means will be excess percentage I th Sou → Al₁₂ (S0₂₁)3 + 3 tho 40kg-mol th 50₂ react in 3 mol In sou will be - 40 kg- mole = 8000 mol 8kg-mol mass of AL₂O3 Mass of 8 kg md Al2(104)3= reaction efficiency = freaction efficiemy T while = = 4014-mal excess mol of the soy Total Mol of the son 40% So remain mole or excess 24 kg-mol 16 Ку-т-1 yoly-md 819-mol A/₂ 0₂ Convert completely. so mixture -*too we know that Abo, is limiting reactant and limiting reactant és completely convest in to product at the end of the reaction, that means 814 Conversion pescantige of Aloz -XIOD 100% 81g. (c) 8-kymol Aboz produce = 8 19- and Ala (sou) by storchrometry of the reaction because 1-mole Abo produce 8y-mol A₂₂03 produced mole x moleculas mass = goromo1 x 1029/² govo mol x 342% / mal mass of Ah₂2 03 (Input) les Mass of Ab(Son) (autoffing 298257 x100= = = 16kg-und Xtoo Imol Al2(SO4)3 8 lg-mol Ak₂ (Sun) } 816 ку 2736 к Al Свси) з 81619 "X1000 27365
Transcribed Image Text:Answer H922 We (A) Al₂O3 + 3H₂SO4 8kg-mol Al₂O3 and mal IMAho react with Mol 1₂2504 8x3 = 24 kg-mol throu we required only 24kg excess reactant is Hesoy and limiting reactant Abo That means above disscasion 8kg-mole react with 24 kg-mol H₂SO We have mole of - excess percentage = of theor Hasou 8 kg-mol Al₂o reacting with have 40 kg-mol H₂₂ Sou that means will be excess percentage I th Sou → Al₁₂ (S0₂₁)3 + 3 tho 40kg-mol th 50₂ react in 3 mol In sou will be - 40 kg- mole = 8000 mol 8kg-mol mass of AL₂O3 Mass of 8 kg md Al2(104)3= reaction efficiency = freaction efficiemy T while = = 4014-mal excess mol of the soy Total Mol of the son 40% So remain mole or excess 24 kg-mol 16 Ку-т-1 yoly-md 819-mol A/₂ 0₂ Convert completely. so mixture -*too we know that Abo, is limiting reactant and limiting reactant és completely convest in to product at the end of the reaction, that means 814 Conversion pescantige of Aloz -XIOD 100% 81g. (c) 8-kymol Aboz produce = 8 19- and Ala (sou) by storchrometry of the reaction because 1-mole Abo produce 8y-mol A₂₂03 produced mole x moleculas mass = goromo1 x 1029/² govo mol x 342% / mal mass of Ah₂2 03 (Input) les Mass of Ab(Son) (autoffing 298257 x100= = = 16kg-und Xtoo Imol Al2(SO4)3 8 lg-mol Ak₂ (Sun) } 816 ку 2736 к Al Свси) з 81619 "X1000 27365
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