oleum

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konspekt zanyatiya primeri resheniya zadach na oleum ximiya s lanoy aksyonovoy 1 primer №1 reshenie navesku 10%-nogo oleuma rastvorili v 100 ml xolodnoy vodi i poluchili 10%- niy rastvor sernoy kisloti. opredelite massu isxodnoy naveski. poskolku oleum soderjit so3, zapishem uravnenie ego reaktsii s vodoy: so3 + h2o = h2so4 pust massa oleuma bila x g, togda massa so3 bila ravna 0,1x g, a sernoy kisloti v oleume bilo 0,9x g. togda zapishem massu konechnogo rastvora: m(h2o isx.) = p•v = 1 • 100 = 100 g m(p-ra konechn.) = m(h2o isx.) + m(oleuma) = 100 + x g srazu mojno virazit massu sernoy kisloti v konechnom rastvore: m(h2so4 konechn.) = m(r-ra konechn.) • w(h2so4)/100% = (100 + x) • 10%/100% = 10 + 0,1x g virazim massu sernoy kisloti, kotoraya obrazovalas iz so3: n(so3) = m(so3)/m(so3) = 0,1x/80 mol n(h2so4 iz so3) = n(so3) = 0,1x/80 mol m(h2so4 iz …

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ri i u mol kisloti. togda virazim massu isxodnoy naveski cherez vvedennie peremennie: m(so3) = n•m = 80x g m(h2so4 iz oleuma) = n•m = 98u g m(naveski) = m(so3) + m(h2so4 iz oleuma) = 80x + 98u g virazim obshee kolichestvo sernoy kisloti, kotoruyu neytralizovali: n(h2so4 iz so3) = n(so3) = x mol n(h2so4 obsh.) = n(h2so4 iz so3) + n(h2so4 iz oleuma)= x + u mol togda mojno zapisat kolichestvo i massu sulfata natriya: n(na2so4) = n(h2so4 obsh.) = x + u mol m(na2so4) = n•m = 142(x + y) g dalee naydem kolichestvo i massu gidroksida natriya: n(naoh) = 2n(na2so4) = 2(x + y) mol m(naoh) = n•m = 80(x + y) g virazim massu isxodnogo rastvora shelochi: m(p-pa naoh) = m(naoh)/w(naoh) • 100% =1600(x + u) g togda massa konechnogo rastvora ravna: m(p-ra konechn.) = m(p-ra naoh) + m(oleuma) = 1600(x + y) + 10,6 g …

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hestvo 30% chast doljna reagirovat so3+h2o → h2so4 m(h2o) = ρ*v = 100 g n(h2o)= = = 5,555 molm 100 m 18 so3 podstavim eti dannie v uravnenie: naydem massu so3 (tu chast, kotoraya proreagirovala) poschitaem massu sernoy kisloti: x1=x2=5,555 mol m(so3)reag. = n*m = 80 g/mol * 5,555 mol = 444,4 g m(h2so4) = n*m = 98 g/mol * 5,555 mol = 544,4 g so3+h2o → h2so4 5,555 molx1 mol x2 mol 1 mol 1 mol 1 mol teper nam nujno poschitat massu so3, kotoriy rastvorilsya: ω(so3)rastv.=30% togda ω(h2so4)=70% a massu sernoy kisloti mi uje znaem! 4 primer №4 kakuyu massu oksida seri (vi) sleduet pribavit k 250 g 60%-noy sernoy kisloti dlya polucheniya 15% oleuma. reshenie 1 2 3 4 5 imeem pravo sostavit proportsiyu :) zdes takje predstavlyaem so3 s dvux pozitsiy dlya resheniya zadachi nam nujno nayti massu kajdoy chasti so3 . nachnem s pervogo. po …

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*30% 70% 8 tot so3 , kotoriy nasitit sernuyu kislotu dlya obrazovaniya samogo 15%-nogo oleuma chast doljna reagirovat so3+h2o → h2so4 so3

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konspekt zanyatiya primeri resheniya zadach na oleum ximiya s lanoy aksyonovoy 1 primer №1 reshenie navesku 10%-nogo oleuma rastvorili v 100 ml xolodnoy vodi i poluchili 10%- niy rastvor sernoy kisloti. opredelite massu isxodnoy naveski. poskolku oleum soderjit so3, zapishem uravnenie ego reaktsii s vodoy: so3 + h2o = h2so4 pust massa oleuma bila x g, togda massa so3 bila ravna 0,1x g, a sernoy kisloti v oleume bilo 0,9x g. togda zapishem massu konechnogo rastvora: m(h2o isx.) = p•v = 1 • 100 = 100 g m(p-ra konechn.) = m(h2o isx.) + m(oleuma) = 100 + x g srazu mojno virazit massu sernoy kisloti v konechnom rastvore: m(h2so4 konechn.) = m(r-ra konechn.) • w(h2so4)/100% = (100 + x) • …

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