{"segment":[{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" H\u00f2a tan 50g mu\u1ed1i \u0103n v\u00e0o 200g n\u01b0\u1edbc thu \u0111\u01b0\u1ee3c dd c\u00f3 n\u1ed3ng \u0111\u1ed9 % l\u00e0","select":["A. 20%. ","B. 20,33%.","C. 30% ","D. 52,7%."],"hint":"","explain":"<span class='basic_left'>$ {{m}_{\\text{dd}}}=50+200=250\\,g \\\\ C{{\\%}_{\\text{dd}}}=\\dfrac{50}{250}.100\\%=20\\% \\\\$ <br\/> <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":1971},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" C\u00e1c oxit axit l\u00e0","select":["A. $CO_2, SiO_2$ ","B. $SO_2, CO$.","C. $P_2O_5, Na_2O$. ","D. $CuO, Fe_2O_3$"],"hint":"","explain":"<span class='basic_left'>Oxit axit l\u00e0 oxit \u0111\u01b0\u1ee3c t\u1ea1o b\u1edfi oxi v\u00e0 1 nguy\u00ean t\u1ed1 phi kim. Oxit axit c\u00f3 th\u1ec3 t\u00e1c d\u1ee5ng v\u1edbi n\u01b0\u1edbc \u0111\u1ec3 t\u1ea1o axit t\u01b0\u01a1ng \u1ee9ng. Ch\u1ec9 c\u00f3 \u0111\u00e1p \u00e1n A th\u1ecfa m\u00e3n<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":1972},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" $Na_2O$ v\u00e0 $Fe_2O_3$ c\u00f9ng ph\u1ea3n \u1ee9ng \u0111\u01b0\u1ee3c v\u1edbi :","select":["A. $H_2O$. ","B. $H_2SO_4$.","C. $NaOH$. ","D. $NaCl$."],"hint":"","explain":"<span class='basic_left'>$Na_2O$ v\u00e0 $Fe_2O_3$ c\u00f9ng l\u00e0 oxit baz\u01a1, c\u00f3 th\u1ec3 ph\u1ea3n \u1ee9ng \u0111\u01b0\u1ee3c v\u1edbi axit.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":1973},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Cho 6,5g Zn v\u00e0o dd HCl th\u00ec th\u1ec3 t\u00edch kh\u00ed $H_2$ tho\u00e1t ra \u1edf \u0111ktc l\u00e0: ","select":["A. 2,24l ","B. 1,12 l","C. 22,4 l ","D. 11,2 l"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& Zn+2HCl\\to ZnC{{l}_{2}}+{{H}_{2}} \\\\ & {{n}_{{{H}_{2}}}}={{n}_{Zn}}=0,1\\,mol \\\\ & {{V}_{{{H}_{2}}}}=2,24\\,l \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":1974},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang.","select":["A. Oxit kim lo\u1ea1i \u0111\u1ec1u l\u00e0 oxit baz\u01a1. ","B. Oxit phi kim \u0111\u1ec1u l\u00e0 oxit baz\u01a1 ","C. C\u00e1c oxit baz\u01a1 \u0111\u1ec1u tan trong n\u01b0\u1edbc t\u1ea1o dung d\u1ecbch baz\u01a1 ","D. N\u01b0\u1edbc v\u00f4i trong l\u00e0m dd phenolphtalein kh\u00f4ng chuy\u1ec3n m\u00e0u."],"hint":"","explain":"<span class='basic_left'>C\u00e1c oxit c\u1ee7a phi kim c\u00f3 th\u1ec3 l\u00e0 oxit trung t\u00ednh. VD: NO, CO <br\/>Ch\u1ec9 c\u00f3 oxit baz\u01a1 c\u1ee7a kim lo\u1ea1i ki\u1ec1m, ki\u1ec1m th\u1ed5 m\u1edbi t\u00e1c d\u1ee5ng v\u1edbi n\u01b0\u1edbc t\u1ea1o th\u00e0nh dung d\u1ecbch baz\u01a1. <br\/>Dung d\u1ecbch n\u01b0\u1edbc v\u00f4i trong $Ca(OH)_2$ c\u00f3 m\u00f4i tr\u01b0\u1eddng ki\u1ec1m n\u00ean l\u00e0m dung d\u1ecbch phenolphtalein chuy\u1ec3n h\u1ed3ng.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":1}]}],"id_ques":1975},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Dug d\u1ecbch l\u00e0m qu\u1ef3 t\u00edm chuy\u1ec3n m\u00e0u \u0111\u1ecf l\u00e0:","select":["A. $HCl, H_2SO_4$.","B. $NaOH, H_2SO_4$.","C. $HCl,H_2O$.","D. $Na_2O, K_2SO_4$."],"hint":"","explain":"<span class='basic_left'>C\u00e1c dung d\u1ecbch axit l\u00e0m qu\u1ef3 chuy\u1ec3n \u0111\u1ecf.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":1976},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Trong c\u00e1c baz\u01a1 sau baz\u01a1 n\u00e0o d\u1ec5 b\u1ecb nhi\u1ec7t ph\u00e2n h\u1ee7y :$KOH, NaOH, Cu(OH)_2, Fe(OH)_2, Fe(OH)_3$.","select":["A. $KOH, Cu(OH)_2, Fe(OH)_3$. ","B. $Cu(OH)_2, Fe(OH)_2, Fe(OH)_3$.","C. $NaOH, Cu(OH)_2, Fe(OH)_2$. ","D. $KOH, NaOH, Fe(OH)_2$."],"hint":"","explain":"<span class='basic_left'>C\u00e1c baz\u01a1 kh\u00f4ng tan b\u1ecb ph\u00e2n h\u1ee7y b\u1edfi nhi\u1ec7t<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":1}]}],"id_ques":1977},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" D\u00e3y ch\u1ea5t n\u00e0o ch\u1ec9 bao g\u1ed3m to\u00e0n mu\u1ed1i:","select":["A. $MgCl_2, Na_2SO_4, KNO_3$. ","B. $Na_2CO_3, H_2SO_4, Ba(OH)_2$","C. $CaSO_4, HCl, MgCO_3$. ","D. $H_2O, Na_3PO_4, KOH$."],"hint":"","explain":"<span class='basic_left'>Mu\u1ed1i l\u00e0 h\u1ee3p ch\u1ea5t c\u1ee7a kim lo\u1ea1i li\u00ean k\u1ebft v\u1edbi g\u1ed1c axit. Ch\u1ec9 c\u00f3 \u0111\u00e1p \u00e1n A th\u1ecfa m\u00e3n<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":1}]}],"id_ques":1978},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Ph\u1ea3n \u1ee9ng c\u1ee7a c\u1eb7p ch\u1ea5t n\u00e0o d\u01b0\u1edbi \u0111\u00e2y kh\u00f4ng t\u1ea1o s\u1ea3n ph\u1ea9m kh\u00ed:","select":["A. Dd $AlCl_3$ + dd $NaOH$.","B. Dd $Al(NO_3)_3$ + dd $Na_2S$. ","C. Dd $AlCl_3$ + dd $Na_2CO_3$. ","D. $Al$ + dd $NaOH$."],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& AlC{{l}_{3}}+3NaOH\\to 3NaCl+Al{{(OH)}_{3}}\\downarrow \\\\ & 2Al{{(N{{O}_{3}})}_{3}}+3N{{a}_{2}}S+6{{H}_{2}}O\\to 6NaN{{O}_{3}}+2Al{{(OH)}_{3}}\\downarrow +3{{H}_{2}}S\\uparrow \\\\ & 2AlC{{l}_{3}}+3N{{a}_{2}}C{{O}_{3}}+3{{H}_{2}}O\\to 6NaCl+2Al{{(OH)}_{3}}+3C{{O}_{2}}\\uparrow \\\\ & 2Al+2NaOH+2{{H}_{2}}O\\to 2NaAl{{O}_{2}}+3{{H}_{2}} \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":1}]}],"id_ques":1979},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Th\u1ec3 t\u00edch dd HCl 2M c\u1ea7n d\u00f9ng \u0111\u1ec3 h\u00f2a tan h\u1ebft 8g CuO l\u00e0:","select":["A. 100ml ","B. 200ml","C. 400ml ","D. 500ml"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& {{n}_{CuO}}=0,1\\,mol \\\\ & CuO+2HCl\\to CuC{{l}_{2}}+{{H}_{2}}O \\\\ & {{n}_{HCl}}=2{{n}_{CuO}}=0,2\\,mol \\\\ & {{V}_{HCl}}=\\dfrac{0,2}{2}=0,1\\,l=100\\,ml \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":1980},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" \u0110\u1ec3 nh\u1eadn bi\u1ebft dd $Na_2SO_4$ v\u00e0 $Na_2CO_3$c\u00f3 th\u1ec3 d\u00f9ng thu\u1ed1c th\u1eed n\u00e0o sau \u0111\u00e2y:","select":["A. Dd $Pb(NO_3)_2$. ","B. Dd $HCl$.","C. Dd $AgNO_3$.","D. Dd $BaCl_2$."],"hint":"","explain":"<span class='basic_left'>Ch\u1ea5t ph\u1ea3n \u1ee9ng v\u1edbi dung d\u1ecbch $HCl$ sinh ra kh\u00ed l\u00e0 $Na_2CO_3$, ch\u1ea5t c\u00f2n l\u1ea1i kh\u00f4ng ph\u1ea3n \u1ee9ng v\u1edbi $HCl$ l\u00e0 $Na_2SO_4$ <br\/>$2HCl+N{{a}_{2}}C{{O}_{3}}\\to 2NaCl+C{{O}_{2}}\\uparrow +{{H}_{2}}O$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":1981},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" C\u1eb7p ch\u1ea5t n\u00e0o sau \u0111\u00e2y c\u00f3 th\u1ec3 c\u00f9ng t\u1ed3n t\u1ea1i trong c\u00f9ng dd:","select":["A. $NaOH$ v\u00e0 $HBr$. ","B. $H_2SO_4$ v\u00e0 $BaCl_2$.","C. $HCl$ v\u00e0 $AgNO_3$. ","D. $NaOH$ v\u00e0 $K_2SO_4$."],"hint":"","explain":"<span class='basic_left'>C\u00e1c ch\u1ea5t c\u00f3 th\u1ec3 t\u1ed3n t\u1ea1i trong 1 dung d\u1ecbch khi ch\u00fang kh\u00f4ng ph\u1ea3n \u1ee9ng v\u1edbi nhau. Ch\u1ec9 c\u00f3 \u0111\u00e1p \u00e1n D th\u1ecfa m\u00e3n.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":1982},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" H\u00f2a tan ho\u00e0n to\u00e0n h\u1ed7n h\u1ee3p X g\u1ed3m 14g (CuO v\u00e0 MgO) v\u00e0o 500ml dd HCl 1M \u0111\u01b0\u1ee3c dung d\u1ecbch A. Kh\u1ed1i l\u01b0\u1ee3ng m gam mu\u1ed1i trong dung d\u1ecbch A l\u00e0","select":["A. 27,75g ","B. 13,5g","C. 15,3g ","D. 25,77g"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& {{n}_{HCl}}=0,5\\,mol \\\\ & CuO+2HCl\\to CuC{{l}_{2}}+{{H}_{2}}O \\\\ & MgO+2HCl\\to MgC{{l}_{2}}+{{H}_{2}}O \\\\ & {{\\text{n}}_{{{H}_{2}}O}}=\\dfrac{1}{2}{{n}_{HCl}}=0,25\\,mol \\\\ & m={{m}_{X}}+{{m}_{HCl}}-{{m}_{{{H}_{2}}O}} \\\\ & \\,\\,\\,\\,\\,\\,\\,=14+0,5.36,5-0,25.18 \\\\ & \\,\\,\\,\\,\\,\\,\\,=27,75\\,gam \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":1983},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Cho thanh $Zn$ v\u00e0o dd $CuSO_4$. Sau 1 tgian l\u1ea5y thanh $Zn$ ra. Bi\u1ebft r\u1eb1ng $Cu$ sinh ra b\u00e1m h\u1ebft v\u00e0o thanh k\u1ebdm th\u00ec kh\u1ed1i l\u01b0\u1ee3ng thanh k\u1ebdm sau p\u1ee9 s\u1ebd:","select":["A. Ko \u0111\u1ed5i. ","B. Gi\u1ea3m xu\u1ed1ng.","C. T\u0103ng l\u00ean.","D. T\u0103ng l\u00ean sau \u0111\u00f3 gi\u1ea3m xu\u1ed1ng."],"hint":"","explain":"<span class='basic_left'>Gi\u1ea3 s\u1eed s\u1ed1 mol Zn ph\u1ea3n \u1ee9ng l\u00e0 x mol<br\/>$\\begin{aligned}& Zn+CuS{{O}_{4}}\\to ZnS{{O}_{4}}+Cu \\\\ & {{m}_{thanh\\,Zn\\text{ }}}={{m}_{Cu}}-{{m}_{\\text{Zn}}} \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=65x-64x=-x \\\\ \\end{aligned}$<br\/>Kh\u1ed1i l\u01b0\u1ee3ng thanh k\u1ebdm gi\u1ea3m.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":1}]}],"id_ques":1984},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Oxit c\u1ee7a m\u1ed9t nguy\u00ean t\u1ed1 M c\u00f3 h\u00f3a tr\u1ecb II ch\u1ee9a 80% M v\u1ec1 kh\u1ed1i l\u01b0\u1ee3ng.Nguy\u00ean t\u1ed1 \u0111\u00f3 l\u00e0","select":["A. Mg. ","B. Fe","C. Cu. ","D. Ca."],"hint":"","explain":"<span class='basic_left'>C\u00f4ng th\u1ee9c c\u1ee7a oxit l\u00e0 MO <br\/>$\\begin{aligned}& \\%{{m}_{O}}=\\dfrac{M}{M+16}.100\\%=80\\% \\\\ & =>M=64\\,(Cu) \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":1985},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Kh\u1eed 12g s\u1eaft (III) oxit b\u1eb1ng kh\u00ed $H_2$. Th\u1ec3 t\u00edch kh\u00ed hi\u0111ro c\u1ea7n d\u00f9ng (\u0111ktc):","select":["A. 22,4 l.","B. 2,24 l.","C. 5,04 l.","D. 1,12 l."],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& {{n}_{F{{e}_{2}}{{O}_{3}}}}=0,075\\,mol \\\\ & F{{e}_{2}}{{O}_{3}}+3{{H}_{2}}\\to 2Fe+3{{H}_{2}}O \\\\ & {{n}_{{{H}_{2}}}}=3{{n}_{F{{e}_{2}}{{O}_{3}}}}=0,225\\,mol \\\\ & {{V}_{{{H}_{2}}}}=5,04\\,l \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":1986},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" C\u00f3 nh\u1eefng kh\u00ed sau $CO_2, H_2, O_2, SO_2, CO$. Kh\u00ed l\u00e0m \u0111\u1ee5c n\u01b0\u1edbc v\u00f4i trong l\u00e0:","select":["A. $CO_2$. ","B. $CO_2, SO_2$.","C. $CO_2, CO, H_2$. ","D. $CO_2, O_2, H_2$."],"hint":"","explain":"<span class='basic_left'>Kh\u00ed l\u00e0m \u0111\u1ee5c n\u01b0\u1edbc v\u00f4i trong: $CO_2, SO_2$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":1987},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" Ph\u00e2n l\u00e2n supe photphat k\u00e9p th\u1ef1c t\u1ebf s\u1ea3n xu\u1ea5t \u0111\u01b0\u1ee3c th\u01b0\u1eddng ch\u1ec9 c\u00f3 40% $P_2O_5$. H\u00e0m l\u01b0\u1ee3ng % c\u1ee7a $Ca(H_2PO_4)_2$ trong ph\u00e2n b\u00f3n \u0111\u00f3","select":["A. 6,59%. ","B. 9,5%.","C. 0,659%.","D. 65,9%."],"hint":"","explain":"<span class='basic_left'>Gi\u1ea3 s\u1eed c\u00f3 100 gam ph\u00e2n b\u00f3n <br\/>$\\begin{aligned}& {{m}_{{{P}_{2}}{{O}_{5}}}}=\\dfrac{40\\%.100}{100\\%}=40\\,gam \\\\ & {{n}_{{{P}_{2}}{{O}_{5}}}}=\\dfrac{20}{71}\\,mol \\\\ & {{n}_{Ca{{({{H}_{2}}P{{O}_{4}})}_{2}}}}={{n}_{{{P}_{2}}{{O}_{5}}}}=\\dfrac{20}{71}\\,mol \\\\ & {{m}_{Ca{{({{H}_{2}}P{{O}_{4}})}_{2}}}}\\approx 65,9\\,gam \\\\ & \\%{{m}_{Ca{{({{H}_{2}}P{{O}_{4}})}_{2}}}}=\\dfrac{65,9}{100}.100\\%=65,9\\% \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":1988},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Cho 6,05g h\u1ed7n h\u1ee3p g\u1ed3m Zn v\u00e0 Fe p\u1ee9 v\u1eeba \u0111\u1ee7 v\u1edbi m g dd HCl 10%. C\u00f4 c\u1ea1n dd sau p\u1ee9 thu \u0111\u01b0\u1ee3c 13,15g mu\u1ed1i khan. Gi\u00e1 tr\u1ecb c\u1ee7a m l\u00e0","select":["A. 73g ","B. 36,6g.","C. 68,4g. ","D. 64g."],"hint":"","explain":"<span class='basic_left'>G\u1ecdi s\u1ed1 mol c\u1ee7a Zn v\u00e0 Fe l\u1ea7n l\u01b0\u1ee3t l\u00e0 x, y <br\/>Kh\u1ed1i l\u01b0\u1ee3ng h\u1ed7n h\u1ee3p b\u1eb1ng 6,05 gam n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: <br\/>65x + 56y = 6,05 (1) <br\/>$\\begin{aligned}& Zn+2HCl\\to ZnC{{l}_{2}}+{{H}_{2}} \\\\ & x\\,\\,\\,x \\\\ & Zn+2HCl\\to ZnC{{l}_{2}}+{{H}_{2}} \\\\ & y\\,\\,\\,y \\\\ \\end{aligned}$ Kh\u1ed1i l\u01b0\u1ee3ng 2 mu\u1ed1i l\u00e0 13,15 gam <br\/>=> 136x + 127y = 13,15 (2)<br\/>Gi\u1ea3i (1) v\u00e0 (2) ta c\u00f3 x = y = 0,05 mol <br\/>$\\begin{aligned}& {{n}_{HCl}}=2x+2y=0,2\\,mol \\\\ & {{m}_{HCl}}=7,3\\,gam \\\\ & {{m}_{\\text{dd}}}=\\dfrac{7,3.100\\%}{10\\%}=73\\,gam \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":1989},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" Kh\u1eed ho\u00e0n to\u00e0n 24g h\u1ed7n h\u1ee3p $CuO$ v\u00e0 $Fe_2O_3$ c\u1ea7n 8,96 l\u00edt kh\u00ed $CO$ \u0111ktc. Th\u00e0nh ph\u1ea7n % m\u1ed7i oxit trong h\u1ed7n h\u1ee3p l\u1ea7n l\u01b0\u1ee3t l\u00e0:","select":["A. 2,08% v\u00e0 97,92%.","B. 26,7% v\u00e0 73,3%. ","C. 20,8% v\u00e0 79,2%. ","D. 33% v\u00e0 67%."],"hint":"","explain":"<span class='basic_left'>G\u1ecdi x, y l\u1ea7n l\u01b0\u1ee3t l\u00e0 s\u1ed1 mol c\u1ee7a CuO v\u00e0 Fe2O3 <br\/>Kh\u1ed1i l\u01b0\u1ee3ng h\u1ed7n h\u1ee3p b\u1eb1ng 40 gam => 80x + 160y = 40 (1)<br\/>$\\begin{aligned}& CuO+CO\\xrightarrow{{{t}^{0}}}Cu+C{{O}_{2}} \\\\ & x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,x \\\\ & F{{e}_{2}}{{O}_{3}}+3CO\\xrightarrow{{{t}^{0}}}2Fe+3C{{O}_{2}} \\\\ & y\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,3y \\\\ & {{n}_{CO}}=0,4\\,mol \\\\ & =>\\,x+3y=0,4(2) \\\\ \\end{aligned}$ <br\/>Gi\u1ea3i (1) v\u00e0 (2) ta c\u00f3 x = y = 0,1 mol <br\/>$\\begin{aligned}& \\%{{m}_{CuO}}=\\dfrac{0,1.80}{24}.100\\%\\approx 33\\% \\\\ & \\%{{m}_{F{{e}_{2}}{{O}_{3}}}}=100\\%-33\\%=67\\% \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":1990}],"lesson":{"save":0,"level":2}}