11. oleum

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oleum erituvchi → sardor voxidov oleum → sulfat angidridning ( 𝑆𝑂3) sulfat kislotadagi eritmasidir. 𝐻2𝑆𝑂4 erigan modda → 𝑆𝑂3 oleum → sulfat angidridning ( 𝑆𝑂3) sulfat kislotadagi eritmasidir. oleum tarkibi 2 xil ifodalanishi mumkin. 1. formula bilan. 𝐻2𝑆𝑂4 ∙ 𝑥𝑆𝑂3 → 𝑥𝐻2 𝑆𝑂4 ∙ 𝑦𝑆𝑂3 x butun son bo’lishi shart emas! 2. foizda ifodalash oleum eritmasi tarkibida 𝑆𝑂3 𝑠𝑢𝑙𝑓𝑎𝑡 𝑎𝑛𝑔𝑖𝑑𝑟𝑖𝑑 miqdori uning foizi hisoblanadi. 20% li oleum→ 𝐻2𝑆𝑂4 + 𝑆𝑂3 80g ____20g____100g(eritma) formula va foiz tarkibni bog’lash. a) formuladan foizga o’tish. 𝐻2𝑆𝑂4 ∙ 1,4𝑆𝑂3 oleumning foiz tarkibini aniqlang. 𝐻2𝑆𝑂4 ∙ 1,4𝑆𝑂3 98g + 112 =210g _____100% 112g ____ x = 53,33% 3𝐻2𝑆𝑂4 ∙ 2𝑆𝑂3 oleumning foiz tarkibini aniqlang. 3𝐻2𝑆𝑂4 ∙ 2𝑆𝑂3 294g + 160 =454g _____100% 160g ____ x = 35,24% sardor voxidov ______ b) foizda berilsa formulani topish. 1. 28,98% li oleumning formulasini aniqlang. oleum 100g 1 – usul (proporsiya) 𝐻2𝑆𝑂4___71,02g 𝑆𝑂3____28,98g 98g____x x = 40g 80 =0,5 2 …

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oleumdan olish mumkin bo’lgan toza kislota massasi. 𝑆𝑂3 +𝐻2𝑂→𝐻2𝑆𝑂4 80g ________98g 100% _______ x = 122,5% → 𝑆𝑂3 𝑠𝑢𝑙𝑓𝑎𝑡 𝑓𝑜𝑖𝑧𝑖. 𝐻2𝑆𝑂4 ∙ 0,5𝑆𝑂3 +0,5𝐻2𝑂→1,5 138g _____________147g 100 _____________ x 106,52 sardor voxidov 𝐻2𝑆𝑂4 = 𝐻2𝑆𝑂4 ∙ 1,4𝑆𝑂3 +1,4𝐻2𝑂→2,4 210g _____________235,2g 100 _____________ x 112% 𝐻2𝑆𝑂4 = 𝐻2𝑆𝑂4 ∙ 2𝑆𝑂3 +2𝐻2𝑂 → 3 258g _____________294g 100 _____________ x 113,95% 𝐻2𝑆𝑂4 = 𝐻2𝑆𝑂4 ∙ 1,25𝑆𝑂3+1,25𝐻2𝑂→ 2,25 198g _____________220,5g 100 _____________ x 111,36% sardor voxidov sulfat foiz = + + 40% li oleumning sulfat foizini aniqlang. 100g____ 𝐻2𝑆𝑂4 60 ____ 𝑆𝑂3 40g 𝑆𝑂3 + 𝐻2𝑂 → 𝐻2𝑆𝑂4 80g __________ 98g 40g __________x = 49g 60g = 49g=109g 100 𝐶%(𝑜𝑙𝑒𝑢𝑚)∙0,225 sulfat foiz = +100 40 ∙ 0,225 109 sardor voxidov + 30% li oleumning sulfat foizini aniqlang. 100g____ 𝐻2𝑆𝑂4 70 ____ 𝑆𝑂3 30g 𝑆𝑂3 + 𝐻2𝑂 → 𝐻2𝑆𝑂4 80g __________ 98g 30g __________x = 36,75g 70g 36,75g= 106,75g sardor voxidov 3. yuduzcha ( oleumning …

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eum tarkibida s va o massa ulshlari farqi 25.69% bo’lgan oleumning 43.6g miqdorini to’liq neytralash uchun qanday massada 32% li naoh eritmasi zarur? javob:125g + 𝐻2𝑆𝑂4 ∙ 𝑥𝑆𝑂3 __________32 + 16x 98 + 80x =0,2569 32 + 16x =25,18 + 20,55x 4,55x = 6,82 x = 1,5 𝐻2𝑆𝑂4 ∙ 1,5𝑆𝑂3 naoh → 𝑁𝑎2𝑆𝑂4 + 𝐻2𝑂2,55 3,5 218g ______ 200g 43,6g ____ x = 40g_____ 0,32 =125g sardor voxidov oleum tarkibida h va o massa ulushlari yig’indisi s massa ulushidan 31,08% ko’p bo’lgan oleumning 39g miqdorini to’liq sulfat kislotaga aylantirish uchun qanday masada suv qo’shish zarur? javob:2,16g + 𝐻2𝑆𝑂4 ∙ 𝑥𝑆𝑂3 __________34 + 16x 98 + 80x =0,3108 34 + 16x =30,46 + 24,86x 8,86x =3,54 x = 0,4 𝐻2𝑆𝑂4 ∙ 0,4𝑆𝑂3 𝐻2𝑂 → 𝐻2𝑆𝑂41,40,4 130g ______ 7,2g 39g ____ x = 2,16g sardor voxidov m(o) 13. oleumda s massa ulushi kislorod massa ulushidan 1,727 marta kichik bo’lsa, shu oleumning foizi …

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avob: 3h 2 so 4 ∙2so 3 𝐻2𝑆𝑂4 ∙ 𝑥𝑆𝑂3 s o ______ = ______ 1 3,6 1 + x 4 + 3x 3,6 + 3,6x = 4 + 3x 0,6x = 0,4 x = 0,667 𝐻2𝑆𝑂4 ∙ 0,667𝑆𝑂3 3𝐻2𝑆𝑂4 ∙ 2𝑆𝑂3 80x = 64 + 48x oleum tarkibida o massa ulushi so 3 massa ulushiga teng bo’lsa. shu oleum formulasi topilsin. javob: h 2 so 4 ∙2so 3𝐻2𝑆𝑂4 ∙ 𝑥𝑆𝑂3 ______ = ______ 1 80x 64 + 48x 1 32x = 64 x = 2 𝐻2𝑆𝑂4 ∙ 2𝑆𝑂3 sardor voxidov 27. oleumda σ va π bog’lar nisbati 2:1 bo’lsa , oleum formulasi topilsin. javob: 3h2so4∙2so3 𝐻2𝑆𝑂4 ∙ 𝑥𝑆𝑂3 σ π ______ = ______ 2 1 6 + 3x 2 + 3x 6 + 3x = 4 + 6x 2 = 3x x = 0,667 𝐻2𝑆𝑂4 ∙ 0,667𝑆𝑂3 3𝐻2𝑆𝑂4 ∙ 2𝑆𝑂3 31. 0.8mol oleum 73g bo’lsa, formulasi topilsin. javob: h2so4∙0.6so3 …

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topilsin. a) 31,36 b) 58,24 c) 28,6 d) 30,2 146g +h 2 so 4 ∙ 0.6so 3 0,6𝐻2𝑂→ 𝐻2𝑆𝑂41,6 ________________ 156,8g 102,2g ___________ x x = 109,76g → hosil bo’lgan kislota 102,2g + 247,8g = 350g _______ 100% 109,76g ____ x x = 31,36% sardor voxidov h2so4∙0.8so3 tarkibli 121,5g oleum, 128,5g suvda eritilganda hosil bo’lgan eritma konsentratsiyasi topilsin. javob: 52,92% 162g +h 2 so 4 ∙ 0.8so 3 0,8𝐻2𝑂→ 𝐻2𝑆𝑂41,8 ________________ 176,4g 121,5g ___________ x x = 132,3g → hosil bo’lgan kislota 121,5g + 128,5g = 250g _______ 100% 132,3g ____ x x = 52,92% sardor voxidov 1 – usul tenglama 44. 120g suvda qanday massada (g) h 2 so 4 ∙1,1so 3 eritilganda hosil bo’lgan eritma 26,18% li bo’ladi? a) 27,9 b) 36,36 c) 54 d) 37.2 h 2 so 4 ∙1,1so 3 205,8x + 1,1 𝐻2𝑂 → 2,1 𝐻2𝑆𝑂4 186x _______________ 205,8x 120g + 186x ______ 100% 205,8x …

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oleum erituvchi → sardor voxidov oleum → sulfat angidridning ( 𝑆𝑂3) sulfat kislotadagi eritmasidir. 𝐻2𝑆𝑂4 erigan modda → 𝑆𝑂3 oleum → sulfat angidridning ( 𝑆𝑂3) sulfat kislotadagi eritmasidir. oleum tarkibi 2 xil ifodalanishi mumkin. 1. formula bilan. 𝐻2𝑆𝑂4 ∙ 𝑥𝑆𝑂3 → 𝑥𝐻2 𝑆𝑂4 ∙ 𝑦𝑆𝑂3 x butun son bo’lishi shart emas! 2. foizda ifodalash oleum eritmasi tarkibida 𝑆𝑂3 𝑠𝑢𝑙𝑓𝑎𝑡 𝑎𝑛𝑔𝑖𝑑𝑟𝑖𝑑 miqdori uning foizi hisoblanadi. 20% li oleum→ 𝐻2𝑆𝑂4 + 𝑆𝑂3 80g ____20g____100g(eritma) formula va foiz tarkibni bog’lash. a) formuladan foizga o’tish. 𝐻2𝑆𝑂4 ∙ 1,4𝑆𝑂3 oleumning foiz tarkibini aniqlang. 𝐻2𝑆𝑂4 ∙ 1,4𝑆𝑂3 98g + 112 =210g _____100% 112g ____ x = 53,33% 3𝐻2𝑆𝑂4 ∙ 2𝑆𝑂3 oleumning foiz tarkibini aniqlang. 3𝐻2𝑆𝑂4 ∙ 2𝑆𝑂3 294g + 160 =454g _____100% 160g ____ x = …

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