{"segment":[{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"Cho 8,4 gam s\u1eaft t\u00e1c d\u1ee5ng v\u1edbi m\u1ed9t l\u01b0\u1ee3ng dung d\u1ecbch HCl v\u1eeba \u0111\u1ee7. D\u1eabn to\u00e0n b\u1ed9 l\u01b0\u1ee3ng kh\u00ed sinh ra qua 16 gam \u0111\u1ed3ng (II) oxit n\u00f3ng. Kh\u1ed1i l\u01b0\u1ee3ng kim lo\u1ea1i \u0111\u1ed3ng thu \u0111\u01b0\u1ee3c sau ph\u1ea3n \u1ee9ng l\u00e0 ","select":["A. 6,4 gam ","B. 9,6 gam ","C. 12 gam ","D. 9 gam "],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{n}_{Fe}}=\\dfrac{8,4}{56}=0,15(mol) \\\\ & {{n}_{CuO}}=\\dfrac{16}{80}=0,2(mol) \\\\ & Fe+2HCl\\to FeC{{l}_{2}}+{{H}_{2}}(1) \\\\ & \\begin{matrix} 0,15 & {} & \\to & {} & {} & 0,15 & {} \\\\\\end{matrix} \\\\ & \\to {{n}_{{{H}_{2}}}}=0,15(mol) \\\\ & CuO+{{H}_{2}}\\xrightarrow{{{t}^{o}}}Cu+{{H}_{2}}O(2) \\\\ \\end{aligned}$<br\/>Theo ph\u01b0\u01a1ng tr\u00ecnh (2) ta c\u00f3 t\u1ef7 s\u1ed1 mol : $\\dfrac{{{n}_{{{H}_{2}}}}}{1}<\\dfrac{{{n}_{CuO}}}{1}$<br\/>$\\to$ Sau ph\u1ea3n \u1ee9ng CuO d\u01b0, ${{H}_{2}}$ h\u1ebft <br\/>L\u01b0\u1ee3ng Cu sinh ra \u0111\u01b0\u1ee3c t\u00ednh theo l\u01b0\u1ee3ng h\u1ebft<br\/>${{n}_{Cu}}={{n}_{{{H}_{2}}}}=0,15(mol)\\to {{m}_{Cu}}=0,15.64=9,6(g)$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2360},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"M\u1ed9t oxit c\u1ee7a \u0111\u1ed3ng c\u00f3 c\u00f4ng th\u1ee9c d\u1ea1ng $C{{u}_{x}}{{O}_{y}}$, bi\u1ebft t\u1ec9 l\u1ec7 kh\u1ed1i l\u01b0\u1ee3ng gi\u1eefa \u0111\u1ed3ng v\u00e0 oxi trong oxit l\u00e0 4:1. Ph\u01b0\u01a1ng tr\u00ecnh \u0111i\u1ec1u ch\u1ebf mu\u1ed1i \u0111\u1ed3ng sunfat t\u1eeb $C{{u}_{x}}{{O}_{y}}$ l\u00e0 ","select":["A. $CuO+{{H}_{2}}S{{O}_{4}}\\to CuS{{O}_{4}}+{{H}_{2}}O$ ","B. $C{{u}_{2}}O+2{{H}_{2}}S{{O}_{4}}\\to 2CuS{{O}_{4}}+{{H}_{2}}O$ ","C. $CuO+{{H}_{2}}S{{O}_{4}}\\to CuS{{O}_{4}}+{{H}_{2}}$ ","D. $CuO+{{H}_{2}}S{{O}_{4}}\\to CuS{{O}_{4}}$ "],"hint":"","explain":"<span class='basic_left'>Kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a Cu v\u00e0 O c\u00f3 trong 1 mol oxit l\u00e0<br\/> $\\begin{aligned} & {{m}_{Cu}}=64x \\\\ & {{m}_{O}}=16y \\\\ \\end{aligned}$<br\/>Theo \u0111\u1ec1 b\u00e0i : <br\/>$\\dfrac{{{m}_{Cu}}}{{{m}_{O}}}=\\dfrac{4}{1}\\to \\dfrac{64x}{16y}=\\dfrac{4}{1}\\to \\dfrac{x}{y}=\\dfrac{1}{1}\\to \\left\\{ \\begin{aligned} & x=1 \\\\ & y=1 \\\\ \\end{aligned} \\right.$<br\/>V\u1eady c\u00f4ng th\u1ee9c ph\u00e2n t\u1eed c\u1ee7a oxit l\u00e0 CuO<br\/>Ph\u01b0\u01a1ng tr\u00ecnh \u0111i\u1ec1u ch\u1ebf mu\u1ed1i \u0111\u1ed3ng sunfat l\u00e0 <br\/>$CuO+{{H}_{2}}S{{O}_{4}}\\to CuS{{O}_{4}}+{{H}_{2}}O$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2361},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"Trong nh\u1eefng ch\u1ea5t d\u01b0\u1edbi \u0111\u00e2y, ch\u1ea5t n\u00e0o l\u00e0m qu\u1ef3 t\u00edm kh\u00f4ng \u0111\u1ed5i m\u00e0u ? ","select":["A. $HN{{O}_{3}}$ ","B. $NaOH$ ","C. $NaCl$ ","D. $Ca{{OH}_{2}}$ "],"hint":"","explain":"<span class='basic_left'>V\u00ec baz\u01a1 l\u00e0m qu\u1ef3 t\u00edm chuy\u1ec3n th\u00e0nh m\u00e0u xanh, axit l\u00e0 qu\u1ef3 t\u00edm chuy\u1ec3n th\u00e0nh m\u00e0u \u0111\u1ecf, n\u00ean ch\u1ec9 c\u00f2n c\u00f3 mu\u1ed1i NaCl kh\u00f4ng l\u00e0m \u0111\u1ed5i m\u00e0u qu\u1ef3 t\u00edm <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2362},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"Cho 2,49 gam h\u1ed7n h\u1ee3p 3 kim lo\u1ea1i Mg, Fe, Zn tan ho\u00e0n to\u00e0n trong dung d\u1ecbch axit sunfuric lo\u00e3ng th\u1ea5y c\u00f3 1,344 l\u00edt kh\u00ed hi\u0111ro tho\u00e1t ra \u1edf \u0111ktc. Kh\u1ed1i l\u01b0\u1ee3ng mu\u1ed1i thu \u0111\u01b0\u1ee3c sau ph\u1ea3n \u1ee9ng l\u00e0 ","select":["A. 13,54 gam ","B. 12,4 gam ","C. 9,54 gam ","D. 8,25 gam "],"hint":"","explain":"<span class='basic_left'>S\u1ed1 mol kh\u00ed ${{H}_{2}}$ tho\u00e1t ra l\u00e0 <br\/>${{n}_{{{H}_{2}}}}=\\dfrac{1,344}{22,4}=0,06(mol)$<br\/>C\u00e1c ph\u01b0\u01a1ng tr\u00ecnh ph\u1ea3n \u1ee9ng x\u1ea3y ra nh\u01b0 sau :<br\/>$\\begin{aligned} & Mg+{{H}_{2}}S{{O}_{4}}\\to MgS{{O}_{4}}+{{H}_{2}} \\\\ & Fe+{{H}_{2}}S{{O}_{4}}\\to FeS{{O}_{4}}+{{H}_{2}} \\\\ & Zn+{{H}_{2}}S{{O}_{4}}\\to ZnS{{O}_{4}}+{{H}_{2}} \\\\ \\end{aligned}$<br\/>Nh\u1eadn x\u00e9t : t\u1eeb c\u00e1c ph\u01b0\u01a1ng tr\u00ecnh ph\u1ea3n \u1ee9ng tr\u00ean ta th\u1ea5y : ${{n}_{{{H}_{2}}S{{O}_{4}}}}={{n}_{{{H}_{2}}}}=0,06(mol)$<br\/>Theo \u0111\u1ecbnh lu\u1eadt b\u1ea3o to\u00e0n kh\u1ed1i l\u01b0\u1ee3ng ta c\u00f3 : <br\/>$\\begin{align} & {{m}_{kl}}+{{m}_{{{H}_{2}}S{{O}_{4}}}}={{m}_{muoi}}+{{m}_{{{H}_{2}}}} \\\\ & \\to {{m}_{muoi}}=2,49+0,06.98-0,06.2=8,25(g) \\\\ \\end{align}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2363},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"C\u00f3 4 ch\u1ea5t r\u1eafn \u1edf d\u1ea1ng b\u1ed9t l\u00e0 $Al,Cu,F{{e}_{2}}{{O}_{3}},CuO$. Thu\u1ed1c th\u1eed n\u00e0o sau \u0111\u00e2y c\u00f3 th\u1ec3 d\u00f9ng \u0111\u1ec3 ph\u00e2n bi\u1ec7t c\u1ea3 4 ch\u1ea5t tr\u00ean ? ","select":["A. $NaOH$ ","B. $HCl$ ","C. Qu\u1ef3 t\u00edm ","D. H\u1ed3 tinh b\u1ed9t "],"hint":"","explain":"<span class='basic_left'>Cho $HCl$ v\u00e0o 4 m\u1eabu th\u1eed l\u00e0 $Al,Cu,F{{e}_{2}}{{O}_{3}},CuO$ \u1edf d\u1ea1ng b\u1ed9t<br\/>M\u1eabu th\u1eed n\u00e0o kh\u00f4ng c\u00f3 ph\u1ea3n \u1ee9ng $\\to$ \u0111\u00f3 l\u00e0 $Cu$<br\/>M\u1eabu th\u1eed n\u00e0o th\u1ea5y c\u00f3 kh\u00ed bay ra $\\to$ \u0111\u00f3 l\u00e0 $Al$<br\/>$2Al+6HCl\\to 2AlC{{l}_{3}}+3{{H}_{2}}\\uparrow $<br\/>M\u1eabu th\u1eed n\u00e0o th\u1ea5y c\u00f3 xu\u1ea5t hi\u1ec7n dung d\u1ecbch m\u00e0u xanh $\\to$ \u0111\u00f3 l\u00e0 $CuO$<br\/>$CuO+2HCl\\to CuC{{l}_{2}}+{{H}_{2}}O$<br\/>M\u00e3u th\u1eed n\u00e0o tan trong dung d\u1ecbch HCl l\u00e0 $F{{e}_{2}}{{O}_{3}}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2364},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"Ng\u01b0\u1eddi ta \u0111i\u1ec7n ph\u00e2n m gam n\u01b0\u1edbc thu \u0111\u01b0\u1ee3c 28 l\u00edt kh\u00ed oxi ( \u0111ktc ). Kh\u1ed1i l\u01b0\u1ee3ng m n\u01b0\u1edbc \u0111\u00e3 b\u1ecb ph\u00e2n h\u1ee7y l\u00e0 ","select":["A. 45 gam ","B. 35 gam ","C. 25 gam ","D. 15 gam "],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{n}_{{{O}_{2}}}}=\\dfrac{28}{22,4}=1,25(mol) \\\\ & 2{{H}_{2}}O\\xrightarrow{dp}2{{H}_{2}}+{{O}_{2}} \\\\ & \\begin{matrix} 2,5mol & \\leftarrow & {} & 1,25mol \\\\\\end{matrix} \\\\ & \\to {{m}_{{{H}_{2}}O}}=2,5.18=45(g) \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2365},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"Cho 0,72 gam Mg t\u00e1c d\u1ee5ng v\u1edbi dung d\u1ecbch ch\u1ee9a 2,92 gam HCl. Kh\u1ed1i l\u01b0\u1ee3ng mu\u1ed1i thu \u0111\u01b0\u1ee3c l\u00e0: ","select":["A. 3,65 gam ","B. 2,85 gam ","C. 4,55 gam ","D. 3,66 gam "],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{n}_{Mg}}=\\dfrac{0,72}{24}=0,03(mol) \\\\ & {{n}_{HCl}}=\\dfrac{2,92}{36,5}=0,08(mol) \\\\ & Mg+2HCl\\to MgC{{l}_{2}}+{{H}_{2}} \\\\ \\end{aligned}$<br\/>L\u1eadp t\u1ef7 s\u1ed1 theo ph\u01b0\u01a1ng tr\u00ecnh : $\\dfrac{{{n}_{Mg}}}{1}<\\dfrac{{{n}_{HCl}}}{2}$<br\/>$\\to$ Sau ph\u1ea3n \u1ee9ng HCl d\u01b0, Mg ph\u1ea3n \u1ee9ng h\u1ebft<br\/>L\u01b0\u1ee3ng mu\u1ed1i sinh ra \u0111\u01b0\u1ee3c t\u00ednh theo l\u01b0\u1ee3ng h\u1ebft <br\/>$\\begin{aligned} & {{n}_{MgC{{l}_{2}}}}={{n}_{Mg}}=0,03(mol) \\\\ & \\to {{m}_{MgC{{l}_{2}}}}=0,03.95=2,85(g) \\\\ \\end{aligned}$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2366},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"Cho 30 gam m\u1ed9t kim lo\u1ea1i R h\u00f3a tr\u1ecb II t\u00e1c d\u1ee5ng h\u1ebft v\u1edbi n\u01b0\u1edbc cho 16,8 l\u00edt kh\u00ed hi\u0111ro (\u0111ktc). Kim lo\u1ea1i R l\u00e0 ","select":["A. Na ","B. K ","C. Ba ","D. Ca "],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{n}_{{{H}_{2}}}}=\\dfrac{16,8}{22,4}=0,75(mol) \\\\ & R+2{{H}_{2}}O\\to R{{(OH)}_{2}}+{{H}_{2}} \\\\ & \\begin{matrix} 0,75mol & \\leftarrow & {} & {} & 0,75mol & {} & {} \\\\\\end{matrix} \\\\ & \\to {{M}_{R}}=\\dfrac{30}{0,75}=40 \\\\ \\end{aligned}$<br\/>V\u1eady R l\u00e0 canxi (Ca)<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2367},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"Ph\u1ea3n \u1ee9ng n\u00e0o sau \u0111\u00e2y l\u00e0 ph\u1ea3n \u1ee9ng h\u00f3a h\u1ee3p ? ","select":["A. $CaO+{{H}_{2}}O\\to Ca{{(OH)}_{2}}$ ","B. $HN{{O}_{3}}+NaOH\\to NaN{{O}_{3}}+{{H}_{2}}O$ ","C. $MgO+2HCl\\to MgC{{l}_{2}}+{{H}_{2}}O$ ","D. $Cu+AgN{{O}_{3}}\\to Cu{{(N{{O}_{3}})}_{2}}+Ag$ "],"hint":"","explain":"<span class='basic_left'>Ph\u1ea3n \u1ee9ng h\u00f3a h\u1ee3p l\u00e0 ph\u1ea3n \u1ee9ng h\u00f3a h\u1ecdc trong \u0111\u00f3 ch\u1ec9 c\u00f3 m\u1ed9t ch\u1ea5t m\u1edbi ( s\u1ea3n ph\u1ea9m ) \u0111\u01b0\u1ee3c t\u1ea1o ra t\u1eeb hai hay nhi\u1ec1u ch\u1ea5t ban \u0111\u1ea7u<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2368},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"Trong d\u00e3y c\u00e1c h\u1ee3p ch\u1ea5t d\u01b0\u1edbi \u0111\u00e2y, d\u00e3y n\u00e0o l\u00e0 mu\u1ed1i trung h\u00f2a ","select":["A. $NaHC{{O}_{3}},NaHS{{O}_{4}},Ca{{(HC{{O}_{3}})}_{2}},KBr$ ","B. $NaHC{{O}_{3}},CuS{{O}_{4}},AgCl,KBr$ ","C. $NaHC{{O}_{3}},BaS{{O}_{4}},KHS{{O}_{4}},KBr$ ","D. $NaCl,CaC{{O}_{3}},CuC{{l}_{2}},AgN{{O}_{3}}$ "],"hint":"","explain":"<span class='basic_left'>Mu\u1ed1i trung h\u00f2a l\u00e0 mu\u1ed1i m\u00e0 trong g\u1ed1c axit kh\u00f4ng c\u00f3 nguy\u00ean t\u1eed hi\u0111ro c\u00f3 th\u1ec3 thay th\u1ec3 b\u1eb1ng nguy\u00ean t\u1eed kim lo\u1ea1i<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2369},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"Cho c\u00e1c oxit sau \u0111\u00e2y : $S{{O}_{2}},PbO,{{K}_{2}}O,BaO,{{N}_{2}}{{O}_{5}},FeO,F{{e}_{2}}{{O}_{3}}$. C\u00f3 bao nhi\u00eau oxit t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi n\u01b0\u1edbc t\u1ea1o baz\u01a1 ","select":["A. 2 ","B. 4 ","C. 6 ","D. 3 "],"hint":"","explain":"<span class='basic_left'>N\u01b0\u1edbc t\u00e1c d\u1ee5ng v\u1edbi m\u1ed9t s\u1ed1 oxit baz\u01a1 ( nh\u01b0 $CaO,CaO,N{{a}_{2}}O,{{K}_{2}}O$ ) t\u1ea1o ra baz\u01a1<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2370},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"Cho 10,4 gam oxit c\u1ee7a m\u1ed9t nguy\u00ean t\u1ed1 kim lo\u1ea1i X h\u00f3a tr\u1ecb II t\u00e1c d\u1ee5ng v\u1edbi dung d\u1ecbch HCl d\u01b0, sau ph\u1ea3n \u1ee9ng t\u1ea1o th\u00e0nh 15,9 gam mu\u1ed1i. Nguy\u00ean t\u1ed1 kim lo\u1ea1i \u0111\u00f3 l\u00e0 ","select":["A. Fe ","B. Cu ","C. Sr ","D. Ca "],"hint":"","explain":"<span class='basic_left'>X c\u00f3 h\u00f3a tr\u1ecb II, Oxi c\u00f3 h\u00f3a tr\u1ecb II<br\/>Theo quy t\u1eafc h\u00f3a tr\u1ecb ta c\u00f3 c\u00f4ng th\u1ee9c oxit d\u1ea1ng : XO<br\/>$\\begin{aligned}& {{n}_{XC{{l}_{2}}}}=\\dfrac{15,9}{{{M}_{X}}+71} \\\\ & XO+2HCl\\to XC{{l}_{2}}+{{H}_{2}}O \\\\ \\end{aligned}$<br\/>Theo ph\u01b0\u01a1ng tr\u00ecnh :<br\/>$\\begin{aligned} & {{n}_{XO}}={{n}_{XC{{l}_{2}}}}=\\dfrac{15,9}{{{M}_{X}}+71} \\\\ & \\to \\dfrac{10,4}{{{M}_{X}}+16}=\\dfrac{15,9}{{{M}_{X}}+71}\\to {{M}_{X}}=88 \\\\ \\end{aligned}$<br\/>V\u1eady X l\u00e0 stronti (Sr)<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2371},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"Cho 4,19 gam h\u1ed7n h\u1ee3p g\u1ed3m hai kim lo\u1ea1i Zn v\u00e0 Fe t\u00e1c d\u1ee5ng h\u1ebft v\u1edbi dung d\u1ecbch HCl. Bi\u1ebft th\u00e0nh ph\u1ea7n $\\%$ kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a Fe trong h\u1ed7n h\u1ee3p l\u00e0 $53,46\\%$. Th\u1ec3 t\u00edch kh\u00ed hi\u0111ro sinh ra \u1edf \u0111ktc l\u00e0 ","select":["A. 1,234 l\u00edt ","B. 1,568 l\u00edt ","C. 2,43 l\u00edt ","D. 1,96 l\u00edt "],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{m}_{Fe}}=\\dfrac{53,46.4,19}{100}=2,24(g)\\to {{n}_{Fe}}=\\dfrac{2,24}{56}=0,04(mol) \\\\ & {{m}_{Zn}}=4,19-2,24=1,95(g)\\to {{n}_{Zn}}=\\dfrac{1,95}{65}=0,03(mol) \\\\ & Zn+2HCl\\to ZnC{{l}_{2}}+{{H}_{2}}(1) \\\\ & \\begin{matrix} 0,03 & \\to & {} & {} & {} & 0,03 & {} \\\\\\end{matrix} \\\\ & Fe+2HCl\\to FeC{{l}_{2}}+{{H}_{2}}(2) \\\\ & \\begin{matrix} 0,04 & \\to & {} & {} & {} & 0,04 & {} \\\\\\end{matrix} \\\\ & \\to {{n}_{{{H}_{2}}}}={{n}_{{{H}_{2}}(1)}}+{{n}_{{{H}_{2}}(2)}}=0,03+0,04=0,07(mol) \\\\ & \\to {{V}_{{{H}_{2}}}}=0,07.22,4=1,568(l) \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2372},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"D\u1eabn 10,08 l\u00edt ( \u1edf \u0111ktc ) h\u1ed7n h\u1ee3p hai kh\u00ed ${{H}_{2}}$ v\u00e0 $CO$ t\u1eeb t\u1eeb qua h\u1ed7n h\u1ee3p hai oxit $FeO$ v\u00e0 $CuO$ nung n\u00f3ng, sau ph\u1ea3n \u1ee9ng th\u1ea5y kh\u1ed1i l\u01b0\u1ee3ng h\u1ed7n h\u1ee3p gi\u1ea3m m gam. Gi\u00e1 tr\u1ecb m l\u00e0 ","select":["A. 11,2 gam ","B. 9,8 gam ","C. 6,5 gam ","D. 7,2 gam "],"hint":"","explain":"<span class='basic_left'>C\u00e1c ph\u01b0\u01a1ng tr\u00ecnh ph\u1ea3n \u1ee9ng l\u00e0 <br\/>$\\begin{aligned} & CuO+CO\\xrightarrow{{{t}^{o}}}Cu+C{{O}_{2}}(1) \\\\ & FeO+CO\\xrightarrow{{{t}^{o}}}Fe+C{{O}_{2}}(2) \\\\ & CuO+{{H}_{2}}\\xrightarrow{{{t}^{o}}}Cu+{{H}_{2}}O(3) \\\\ & FeO+{{H}_{2}}\\xrightarrow{{{t}^{o}}}Fe+{{H}_{2}}O(4) \\\\ \\end{aligned}$<br\/>Theo c\u00e1c ph\u01b0\u01a1ng tr\u00ecnh h\u00f3a h\u1ecdc (1), (2), (3), (4) s\u1ed1 mol nguy\u00ean t\u1eed oxi trong oxit m\u1ea5t \u0111i b\u1eb1ng s\u1ed1 mol CO hay ${{H}_{2}}$<br\/>$\\to$ Kh\u1ed1i l\u01b0\u1ee3ng ch\u1ea5t r\u1eafn gi\u1ea3m : $\\dfrac{10,08}{22,4}.16=7,2(g)$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2373},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"C\u00f4ng th\u1ee9c h\u00f3a h\u1ecdc c\u1ee7a mu\u1ed1i c\u00f3 t\u00ean g\u1ecdi l\u00e0 magie hi\u0111rocacbonat ?","select":["A. $MgHC{{O}_{3}}$ ","B. $MgC{{O}_{3}}$ ","C. $Mg{{(HC{{O}_{3}})}_{2}}$ ","D. $Mg{{(C{{O}_{3}})}_{2}}$ "],"hint":"","explain":"<span class='basic_left'>C\u00f4ng th\u1ee9c h\u00f3a h\u1ecdc c\u1ee7a mu\u1ed1i g\u1ed3m 2 ph\u1ea7n : kim lo\u1ea1i v\u00e0 g\u1ed1c axit<br\/>Kim lo\u1ea1i l\u00e0 Magie (Mg) h\u00f3a tr\u1ecb II<br\/>G\u1ed1c axit l\u00e0 $-HC{{O}_{3}}$ h\u00f3a tr\u1ecb I<br>Theo quy t\u1eafc h\u00f3a tr\u1ecb : $Mg{{(HC{{O}_{3}})}_{2}}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2374},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"Mu\u1ed1i n\u00e0o trong c\u00e1c mu\u1ed1i sau kim lo\u1ea1i c\u00f3 h\u00f3a tr\u1ecb II : $A{{l}_{2}}{{(S{{O}_{4}})}_{3}},NaCl,MgC{{O}_{3}},FeC{{l}_{2}},AgN{{O}_{3}}$ ? ","select":["A. $FeC{{l}_{2}},AgN{{O}_{3}}$ ","B. $MgC{{O}_{3}},FeC{{l}_{2}}$ ","C. $MgC{{O}_{3}}$ ","D. $NaCl,MgC{{O}_{3}}$ "],"hint":"","explain":"<span class='basic_left'>G\u1ed1c $=S{{O}_{4}}$ h\u00f3a tr\u1ecb II<br\/>G\u1ed1c $-Cl$ h\u00f3a tr\u1ecb I<br\/>G\u1ed1c $=C{{O}_{3}}$ h\u00f3a tr\u1ecb II<br\/>$-N{{O}_{3}}$ h\u00f3a tr\u1ecb I<br\/>Theo quy t\u1eafc tr\u1ecb : Al h\u00f3a tr\u1ecb III, Na h\u00f3a tr\u1ecb I, Mg h\u00f3a tr\u1ecb II, Fe h\u00f3a tr\u1ecb II, Ag h\u00f3a tr\u1ecb I<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2375},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"Cho 6,2 gam $N{{a}_{2}}O$ v\u00e0o n\u01b0\u1edbc. Kh\u1ed1i l\u01b0\u1ee3ng NaOH thu \u0111\u01b0\u1ee3c l\u00e0 bao nhi\u00eau ? ","select":["A. 10 gam ","B. 9 gam ","C. 8,6 gam ","D. 8 gam "],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{n}_{N{{a}_{2}}O}}=\\dfrac{6,2}{23.2+16}=0,1(mol) \\\\ & N{{a}_{2}}O+{{H}_{2}}O\\to 2NaOH \\\\ & \\begin{matrix} 0,1mol & \\to & {} & 0,2mol \\\\\\end{matrix} \\\\ & \\to {{m}_{NaOH}}=0,2.40=8(g) \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2376},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"M\u1ed9t s\u1ed1 h\u00f3a ch\u1ea5t \u0111\u01b0\u1ee3c \u0111\u1ec3 tr\u00ean 1 ng\u0103n t\u1ee7 c\u00f3 khung b\u1eb1ng kim lo\u1ea1i. Sau m\u1ed9t n\u0103m ng\u01b0\u1eddi ta th\u1ea5y khung kim lo\u1ea1i b\u1ecb g\u1ec9. H\u00f3a ch\u1ea5t n\u00e0o sau \u0111\u00e2y c\u00f3 kh\u1ea3 n\u0103ng g\u00e2y ra hi\u1ec7n t\u01b0\u1ee3ng tr\u00ean ? ","select":["A. Axit clohi\u0111ric ","B. D\u1ea7u h\u1ecfa ","C. D\u00e2y nh\u00f4m ","D. R\u01b0\u1ee3u etylic (etanol) "],"hint":"","explain":"<span class='basic_left'>Axit l\u00e0 h\u1ee3p ch\u1ea5t h\u00f3a h\u1ecdc c\u00f3 t\u00ednh \u0103n m\u00f2n cao. Khi m\u1ed9t s\u1ed1 kim lo\u1ea1i ti\u1ebfp x\u00fac v\u1edbi axit, th\u00ec axit l\u00e0 t\u00e1c nh\u00e2n oxi h\u00f3a kim lo\u1ea1i th\u00e0nh mu\u1ed1i v\u00e0 kh\u00ed hi\u0111ro<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2377},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"Kh\u1eed ho\u00e0n to\u00e0n 2,4 gam h\u1ed7n h\u1ee3p $CuO$ v\u00e0 $F{{e}_{x}}{{O}_{y}}$ c\u00f3 c\u00f9ng s\u1ed1 mol nh\u01b0 nhau b\u1eb1ng hi\u0111ro \u0111\u01b0\u1ee3c 1,76 gam kim lo\u1ea1i. H\u00f2a tan kim lo\u1ea1i \u0111\u00f3 b\u1eb1ng dung d\u1ecbch HCl d\u01b0 th\u1ea5y tho\u00e1t ra 0,448 l\u00edt ${{H}_{2}}$ (\u0111ktc). C\u00f4ng th\u1ee9c c\u1ee7a oxit s\u1eaft l\u00e0 ","select":["A. $F{{e}_{3}}{{O}_{4}}$ ","B. $FeO$ ","C. $F{{e}_{2}}{{O}_{3}}$ ","D. $F{{e}_{3}}{{O}_{2}}$ "],"hint":"","explain":"<span class='basic_left'>Ph\u01b0\u01a1ng tr\u00ecnh kh\u1eed h\u1ed7n h\u1ee3p oxit :<br\/>$\\begin{align} & CuO+{{H}_{2}}\\xrightarrow{{{t}^{o}}}Cu+{{H}_{2}}O(1) \\\\ & F{{e}_{x}}{{O}_{y}}+y{{H}_{2}}\\xrightarrow{{{t}^{o}}}xFe+y{{H}_{2}}O(2) \\\\ \\end{align}$<br\/>Kim lo\u1ea1i thu \u0111\u01b0\u1ee3c g\u1ed3m Cu v\u00e0 Fe, khi cho h\u1ed7n h\u1ee3p kim lo\u1ea1i n\u00e0y v\u00e0o axit HCl ch\u1ec9 c\u00f3 Fe tham gia ph\u1ea3n \u1ee9ng v\u00ec Cu kh\u00f4ng t\u00e1c d\u1ee5ng v\u1edbi HCl<br\/>$\\begin{aligned} & {{n}_{{{H}_{2}}}}=\\dfrac{0,448}{22,4}=0,02(mol) \\\\ & Fe+2HCl\\to FeC{{l}_{2}}+{{H}_{2}} \\\\ & \\begin{matrix} 0,02mol & \\leftarrow & {} & {} & 0,02mol & {} & {} \\\\\\end{matrix} \\\\ & \\to {{m}_{Fe}}=0,02.56=1,12(g) \\\\ \\end{aligned}$<br\/>Theo b\u00e0i ra kh\u1ed1i l\u01b0\u1ee3ng kim lo\u1ea1i thu \u0111\u01b0\u1ee3c khi kh\u1eed oxit b\u1eb1ng hi\u0111ro l\u00e0 1,76 gam n\u00ean <br\/>$\\begin{aligned} & \\to {{m}_{Cu}}=1,76-1,12=0,64(g) \\\\ & \\to {{n}_{Cu}}=\\frac{0,64}{64}=0,01(mol) \\\\ \\end{aligned}$<br\/>Theo ph\u01b0\u01a1ng tr\u00ecnh (1) : <br\/>$\\begin{aligned} & {{n}_{CuO}}={{n}_{Cu}}=0,01(mol) \\\\ & \\to {{m}_{CuO}}=0,01.80=0,8(g) \\\\ \\end{aligned}$<br\/>V\u1eady kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a $F{{e}_{x}}{{O}_{y}}$ l\u00e0 2,4 - 0,8 = 1,6 gam<br\/>Theo ph\u01b0\u01a1ng tr\u00ecnh (2) :<br\/>$\\begin{aligned} & {{n}_{F{{e}_{x}}{{O}_{y}}}}=\\dfrac{{{n}_{Fe}}}{x}=\\dfrac{0,02}{x} \\\\ & \\to {{M}_{F{{e}_{x}}{{O}_{y}}}}=\\dfrac{1,6x}{0,02}=80x \\\\ & \\to 56x+16y=80x \\\\ & \\to \\dfrac{x}{y}=\\dfrac{2}{3} \\\\ \\end{aligned}$<br\/>V\u1eady c\u00f4ng th\u1ee9c c\u1ee7a oxit s\u1eaft l\u00e0 $F{{e}_{2}}{{O}_{3}}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2378},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"M\u1ed9t ch\u1ea5t l\u1ecfng kh\u00f4ng m\u00e0u c\u00f3 kh\u1ea3 n\u0103ng h\u00f3a \u0111\u1ecf m\u1ed9t ch\u1ea5t ch\u1ec9 th\u1ecb th\u00f4ng d\u1ee5ng. N\u00f3 t\u00e1c d\u1ee5ng v\u1edbi m\u1ed9t s\u1ed1 kim lo\u1ea1i gi\u1ea3i ph\u00f3ng kh\u00ed hi\u0111ro v\u00e0 n\u00f3 gi\u1ea3i ph\u00f3ng kh\u00ed cacbonic khi th\u00eam v\u00e0o mu\u1ed1i hi\u0111rocacbonat. K\u1ebft lu\u1eadn n\u00e0o sau \u0111\u00e2y ph\u00f9 h\u1ee3p nh\u1ea5t cho ch\u1ea5t l\u1ecfng ban \u0111\u1ea7u ? ","select":["A. Ch\u1ea5t l\u1ecfng l\u00e0 mu\u1ed1i ","B. Ch\u1ea5t l\u1ecfng l\u00e0 baz\u01a1 ","C. Ch\u1ea5t l\u1ecfng l\u00e0 axit ","D. Ch\u1ea5t l\u1ecfng l\u00e0 x\u00fat "],"hint":"","explain":"<span class='basic_left'>Ch\u1ea5t l\u1ecfng \u0111\u00f3 l\u00e0 axit<br\/>+ L\u00e0m qu\u1ef3 t\u00edm chuy\u1ec3n th\u00e0nh \u0111\u1ecf<br\/>+ T\u00e1c d\u1ee5ng v\u1edbi kim lo\u1ea1i gi\u1ea3i ph\u00f3ng kh\u00ed hi\u0111ro<br\/>VD : $2Al+6HCl\\to 2AlC{{l}_{3}}+3{{H}_{2}}$<br\/>+ T\u00e1c d\u1ee5ng v\u1edbi mu\u1ed1i hi\u0111rocacbonat gi\u1ea3i ph\u00f3ng kh\u00ed cacbonic<br\/>VD : $HCl+NaHC{{O}_{3}}\\to NaCl+C{{O}_{2}}+{{H}_{2}}O$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2379}],"lesson":{"save":0,"level":2}}