\n
B\u00e0i 2 (trang 157 sgk H\u00f3a 12 n\u00e2ng cao):\u00a0Trong qu\u00e1 tr\u00ecnh \u0111i\u1ec7n ph\u00e2n dung d\u1ecbch KBr, ph\u1ea3n \u1ee9ng n\u00e0o sau \u0111\u00e2y x\u1ea3y ra \u1edf c\u1ef1c d\u01b0\u01a1ng (anot)?<\/strong><\/p>\nA. Ion Br–<\/sup>\u00a0b\u1ecb oxi h\u00f3a<\/p>\nB. Ion Br–<\/sup>\u00a0b\u1ecb kh\u1eed<\/p>\nC. Ion K+<\/sup>\u00a0b\u1ecb oxi h\u00f3a<\/p>\nD. Ion K+<\/sup>\u00a0b\u1ecb kh\u1eed<\/p>\nL\u1eddi gi\u1ea3i:<\/b><\/p>\n
\u0110\u00e1p \u00e1n A<\/p>\n
B\u00e0i 3 (trang 157 sgk H\u00f3a 12 n\u00e2ng cao):\u00a0Cho 3,1 gam h\u1ed7n h\u1ee3p hai kim lo\u1ea1i ki\u1ec1m \u1edf hai chu k\u1ef3 li\u00ean ti\u1ebfp trong b\u1ea3ng tu\u1ea7n ho\u00e0n t\u00e1c d\u1ee5ng h\u1ebft v\u1edbi n\u01b0\u1edbc thu \u0111\u01b0\u1ee3c 1,12 l\u00edt kh\u00ed H2<\/sub>(\u0111ktc)<\/strong><\/p>\na) X\u00e1c \u0111\u1ecbnh t\u00ean hai kim lo\u1ea1i ki\u1ec1m v\u00e0 t\u00ednh % kh\u1ed1i l\u01b0\u1ee3ng m\u1ed7i kim lo\u1ea1i trong h\u1ed7n h\u1ee3p.<\/strong><\/p>\nb) T\u00ednh th\u1ec3 t\u00edch dung d\u1ecbch HCl c\u1ea7n d\u00f9ng \u0111\u1ec3 trung h\u00f2a dung d\u1ecbch ki\u1ec1m tr\u00ean v\u00e0 kh\u1ed1i l\u01b0\u1ee3ng mu\u1ed1i clorua thu \u0111\u01b0\u1ee3c<\/strong><\/p>\nL\u1eddi gi\u1ea3i:<\/b><\/p>\n
\u0110\u1eb7t c\u00f4ng th\u1ee9c chung cho 2 kim lo\u1ea1i ki\u1ec1m l\u00e0 M<\/p>\n
M + H2<\/sub>O \u2192 MOH + 1\/2H2<\/sub><\/p>\nnM<\/sub>\u00a0= 2nH2<\/sub>\u00a0= 2 x (1,12 : 22,4) = 0,1 mol => MM<\/sub>\u00a0= 3,1 : 0,1 = 31<\/p>\na)V\u1eady hai kim lo\u1ea1i ki\u1ec1m li\u00ean ti\u1ebfp l\u00e0 Na(23) v\u00e0 K(39)<\/p>\n
Theo s\u01a1 \u0111\u1ed3 \u0111\u01b0\u1eddng ch\u00e9o<\/p>\n
![\"\"](\"https:\/\/lop12.edu.vn\/wp-content\/uploads\/2018\/04\/bai-3-trang-157-sgk-hoa-12-nang-cao.png\")
<\/p>\n
=> %(m)Na = (0,05 x 23):3,1 x100% = 37,1%<\/p>\n
=> %(m)K = 100 \u2013 37,1 = 62,9%<\/p>\n
b) Ph\u1ea3n \u1ee9ng trung h\u00f2a<\/p>\n
MOH + HCl \u2192 MCl + H2<\/sub>O<\/p>\nnHCl<\/sub>\u00a0= nMOH<\/sub>\u00a0= 0,1 mol => VHCl2M<\/sub>\u00a0= 0,1 : 2 = 0,05 l\u00edt = 50 ml<\/p>\nKh\u1ed1i l\u01b0\u1ee3ng mu\u1ed1i : mMCl<\/sub>\u00a0= 0,1(M + 35,5) = 6,65 gam<\/p>\nB\u00e0i 4 (trang 157 sgk H\u00f3a 12 n\u00e2ng cao):\u00a0Cho 3,9 gam kim lo\u1ea1i k t\u00e1c dung v\u1edbi 101,8 gam n\u01b0\u1edbc, T\u00ednh n\u1ed3ng \u0111\u1ed9 mol v\u00e0 n\u1ed3ng \u0111\u1ed9 % c\u1ee7a ch\u1ea5t trong dung d\u1ecbch thu \u0111\u01b0\u1ee3c. Bi\u1ebft kh\u1ed1i l\u01b0\u1ee3ng ri\u00eang c\u1ee7a dung d\u1ecbch \u0111\u00f3 l\u00e0 1,056 g\/ml<\/strong><\/p>\nL\u1eddi gi\u1ea3i:<\/b><\/p>\n
nK<\/sub>\u00a0= 3,9 : 39 = 0,1 mol<\/p>\nPh\u1ea3n \u1ee9ng : 2K + 2H2<\/sub>O \u2192 2KOH + H2<\/sub><\/p>\nnH2<\/sub>\u00a0= 0,05.2 = 0,1 gam<\/p>\nKh\u1ed1i l\u01b0\u1ee3ng dung d\u1ecbch sau ph\u1ea3n \u1ee9ng :<\/p>\n
mdd<\/sub>\u00a0= 3,9 + 101,8 \u2013 0,1 = 105,6 gam<\/p>\nC%KOH = (0,1.56) : 105,6 . 100 = 5,3%<\/p>\n
Vdd<\/sub>\u00a0= mdd<\/sub>\u00a0: D = 105,6 : 1,056 = 100ml = 0,1 l\u00edt<\/p>\nCMKOH<\/sub>\u00a0= 0,1 : 0,1 = 1M<\/p>\nB\u00e0i 5 (trang 157 sgk H\u00f3a 12 n\u00e2ng cao):\u00a0So s\u00e1nh t\u00ednh ch\u1ea5t h\u00f3a h\u1ecdc c\u1ee7a hai mu\u1ed1i NaHCO3<\/sub>\u00a0v\u00e0 Na2<\/sub>CO3<\/sub>\u00a0Vi\u1ebft c\u00e1c ph\u01b0\u01a1ng tr\u00ecnh h\u00f3a h\u1ecdc minh h\u1ecda.<\/strong><\/p>\nL\u1eddi gi\u1ea3i:<\/b><\/p>\n
\u0110\u1ec1u l\u00e0 mu\u1ed1i c\u1ee7a axit y\u1ebfu, c\u00f3 kh\u1ea3 n\u0103ng nh\u1eadn proton th\u1ec3 hi\u1ec7n t\u00ednh bazo:<\/p>\n
Na2<\/sub>CO3<\/sub>\u00a0+ 2HCl \u2192 2NaCl + CO2<\/sub>\u00a0+ H2<\/sub>O<\/p>\nCO3<\/sub>2-<\/sup>\u00a0+ 2H+<\/sub>\u00a0\u2192 CO2<\/sub>\u00a0+ H2<\/sub>O<\/p>\nNaHCO3<\/sub>\u00a0+ HCl \u2192 NaCl + CO2<\/sub>\u00a0+ H2<\/sub>O<\/p>\nHCO3<\/sub>–<\/sup>\u00a0+ H+ \u2192 CO2<\/sub>\u00a0+ H2<\/sub>O<\/p>\n– NaHCO3<\/sub>\u00a0l\u00e0 mu\u1ed1i axit t\u00e1c d\u1ee5ng v\u1edbi bazo t\u1ea1o mu\u1ed1i trung h\u00f2a<\/p>\nNaHCO3<\/sub>\u00a0+ NaOH \u2192 Na2<\/sub>CO3<\/sub>\u00a0+ H2<\/sub>O<\/p>\nHCO3<\/sub>–<\/sup>\u00a0+ OH–<\/sup>\u00a0\u2192 CO3<\/sub>2-<\/sup>\u00a0+ H2<\/sub>O<\/p>\nV\u1eady NaHCO3<\/sub>\u00a0l\u00e0 mu\u1ed1i c\u00f3 t\u00ednh l\u01b0\u1ee1ng t\u00ednh<\/p>\nNa2<\/sub>CO3<\/sub>\u00a0l\u00e0 mu\u1ed1i c\u00f3 t\u00ednh ch\u1ea5t baz\u01a1<\/p>\nB\u00e0i 6 (trang 157 sgk H\u00f3a 12 n\u00e2ng cao):\u00a0Nung 4,84g h\u1ed7n h\u1ee3p NaHCO3<\/sub>\u00a0v\u00e0 KHCO3<\/sub>\u00a0\u0111\u1ebfn ph\u1ea3n \u1ee9ng ho\u00e0n to\u00e0n thu \u0111\u01b0\u1ee3c 0,56 l\u00edt CO2<\/sub>\u00a0(\u0111ktc). X\u00e1c \u0111\u1ecbnh kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a m\u1ed7i mu\u1ed1i trong h\u1ed7n h\u1ee3p tr\u01b0\u1edbc v\u00e0 sau khi nung.<\/strong><\/p>\nL\u1eddi gi\u1ea3i:<\/b><\/p>\n
<\/p>\n
\u0110\u1eb7t s\u1ed1 mol NaHCO3<\/sub>\u00a0v\u00e0 KHCO3<\/sub>\u00a0l\u00e0 x, y<\/p>\n=> 84x + 100y = 4,84 (1)<\/p>\n
2KHCO3<\/sub>\u00a0to<\/sup>\u00a0\u2192 FCO3<\/sub>\u00a0+ CO2<\/sub>\u00a0+ H2<\/sub>O<\/p>\n2NaHCO3<\/sub>\u00a0to<\/sup>\u00a0\u2192 Na2<\/sub>CO3<\/sub>\u00a0+ CO2<\/sub>\u00a0+ H2<\/sub>O<\/p>\nnCO2<\/sub>\u00a0= (x + y) : 2 = 0,56 : 22,4 => x + y = 0,05 (2) T\u1eeb (1, 2) => x = 0,01; y = 0,04<\/p>\nV\u1eady kh\u1ed1i l\u01b0\u1ee3ng NaHCO3<\/sub>\u00a0l\u00e0 0,01.84 = 0,84 gam<\/p>\nKHCO3<\/sub>\u00a0l\u00e0 0,04.100 = 4,00 gam<\/p>\n\n
<\/div>\n<\/div>\n<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"
B\u00e0i 1 (trang 157 sgk H\u00f3a 12 n\u00e2ng cao):\u00a0Trong qu\u00e1 tr\u00ecnh \u0111i\u1ec7n ph\u00e2n dung d\u1ecbch NaCl, \u1edf c\u1ef1c \u00e2m x\u1ea3y ra : A. S\u1ef1 kh\u1eed ion Na+ B. S\u1ef1 oxi h\u00f3a Na+ C. S\u1ef1 kh\u1eed ph\u00e2n t\u1eed H2O D. S\u1ef1 oxi h\u00f3a ph\u00e2n t\u1eed H2O L\u1eddi gi\u1ea3i: \u0110\u00e1p \u00e1n C B\u00e0i 2 (trang 157 sgk […]<\/p>\n","protected":false},"author":3,"featured_media":25181,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1306],"tags":[1436,1435],"yoast_head":"\n
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