{"segment":[{"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 $Cu$ t\u00e1c d\u1ee5ng v\u1edbi dung d\u1ecbch axit $HCl$ lo\u00e3ng s\u1ebd c\u00f3 hi\u1ec7n t\u01b0\u1ee3ng n\u00e0o sau \u0111\u00e2y? ","select":["A. Ch\u1ea5t kh\u00ed ch\u00e1y \u0111\u01b0\u1ee3c trong kh\u00f4ng kh\u00ed v\u1edbi ngon l\u1eeda m\u00e0u xanh ","B. Ch\u1ea5t kh\u00ed l\u00e0m \u0111\u1ee5c n\u01b0\u1edbc v\u00f4i trong ","C. Dung d\u1ecbch c\u00f3 m\u00e0u xanh ","D. Kh\u00f4ng c\u00f3 hi\u1ec7n t\u01b0\u1ee3ng g\u00ec "],"hint":"","explain":"<span class='basic_left'>Trong ph\u00f2ng th\u00ed nghi\u1ec7m, kh\u00ed Hi\u0111ro \u0111\u01b0\u1ee3c \u0111i\u1ec1u ch\u1ebf b\u1eb1ng c\u00e1ch cho axit $(HCl,{{H}_{2}}S{{O}_{4}})$ lo\u00e3ng t\u00e1c d\u1ee5ng v\u1edbi kim lo\u1ea1i k\u1ebdm (ho\u1eb7c s\u1eaft, nh\u00f4m) c\u00f2n c\u00e1c kim lo\u1ea1i y\u1ebfu nh\u01b0 $Cu, Ag$... kh\u00f4ng c\u00f3 ph\u1ea3n \u1ee9ng v\u1edbi $(HCl,{{H}_{2}}S{{O}_{4}})$ lo\u00e3ng<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":1}]}],"id_ques":2300},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"\u1ee8ng d\u1ee5ng c\u1ee7a Hi\u0111ro l\u00e0: ","select":["A. Oxi h\u00f3a kim lo\u1ea1i ","B. L\u00e0m nguy\u00ean li\u1ec7u cho s\u1ea3n xu\u1ea5t $N{{H}_{3}},HCl$ v\u00e0 ch\u1ea5t h\u1eefu c\u01a1 ","C. T\u1ea1o hi\u1ec7u \u1ee9ng nh\u00e0 k\u00ednh ","D. T\u1ea1o m\u01b0a axit "],"hint":"","explain":"<span class='basic_left'>\u1ee8ng d\u1ee5ng c\u1ee7a hi\u0111ro l\u00e0 h\u00e0n c\u1eaft kim lo\u1ea1i, n\u1ea1p v\u00e0o kh\u00ed c\u1ea7u, kh\u1eed oxi c\u1ee7a m\u1ed9t s\u1ed1 oxit kim lo\u1ea1i, s\u1ea3n xu\u1ea5t amoniac, ph\u00e2n \u0111\u1ea1m, s\u1ea3n xu\u1ea5t nhi\u00ean li\u1ec7u, s\u1ea3n xu\u1ea5t axit clohidric.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":1}]}],"id_ques":2301},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"Kh\u00ed n\u00e0o l\u00e0 kh\u00ed nh\u1eb9 nh\u1ea5t trong c\u00e1c kh\u00ed sau: ","select":["A. ${{H}_{2}}$ ","B. $C{{O}_{2}}$ ","C. ${{H}_{2}}O$ ","D. ${{O}_{2}}$ "],"hint":"","explain":"<span class='basic_left'>Kh\u00ed hi\u0111ro l\u00e0 kh\u00ed nh\u1eb9 nh\u1ea5t trong t\u1ea5t c\u1ea3 c\u00e1c kh\u00ed<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2302},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"Ti l\u1ec7 mol c\u1ee7a Hi\u0111ro v\u1edbi Oxi s\u1ebd g\u00e2y n\u1ed5 m\u1ea1nh l\u00e0: ","select":["A. $1:3$ ","B. $1:1$ ","C. $2:1$ ","D. $1:2$ "],"hint":"","explain":"<span class='basic_left'>Ph\u1ea3n \u1ee9ng gi\u1eefa Hi\u0111ro v\u00e0 Oxi l\u00e0 $2{{H}_{2}}+{{O}_{2}}\\xrightarrow{{{t}^{o}}}2{{H}_{2}}O$<br\/>Theo ph\u01b0\u01a1ng tr\u00ecnh c\u1ee9 $2$ mol Hi\u0111ro s\u1ebd tham gia ph\u1ea3n \u1ee9ng v\u1edbi $1$ mol Oxi<br\/>$\\to$ t\u1ef7 l\u1ec7 mol c\u1ee7a Hi\u0111ro v\u1edbi Oxi l\u00e0 $2:1$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2303},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"Oxit n\u00e0o sau \u0111\u00e2y b\u1ecb kh\u1eed b\u1edfi Hi\u0111ro: ","select":["A. $N{{a}_{2}}O$ ","B. $F{{e}_{2}}{{O}_{3}}$ ","C. $CaO$ ","D. $BaO$ "],"hint":"","explain":"<span class='basic_left'>${{H}_{2}}$ ch\u1ec9 c\u00f3 kh\u1ea3 n\u0103ng kh\u1eed oxit c\u1ee7a kim lo\u1ea1i trung b\u00ecnh v\u00e0 y\u1ebfu (t\u1eeb $Zn$ tr\u1edf xu\u1ed1ng)<br\/>${{H}_{2}}$ kh\u00f4ng kh\u1eed \u0111\u01b0\u1ee3c oxit c\u1ee7a kim lo\u1ea1i m\u1ea1nh nh\u01b0 $MgO, A{{l}_{2}}{{O}_{3}},N{{a}_{2}}O,BaO,CaO$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2304},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"T\u00ean g\u1ecdi kh\u00e1c c\u1ee7a ch\u1ea5t kh\u1eed l\u00e0: ","select":["A. Ch\u1ea5t oxi h\u00f3a ","B. Ch\u1ea5t b\u1ecb oxi h\u00f3a ","C. Ch\u1ea5t b\u1ecb kh\u1eed ","D. Ch\u1ea5t l\u1ea5y oxi "],"hint":"","explain":"<span class='basic_left'>Ch\u1ea5t kh\u1eed l\u00e0 ch\u1ea5t chi\u1ebfm oxi c\u1ee7a ch\u1ea5t kh\u00e1c v\u00e0 s\u1ef1 oxi h\u00f3a l\u00e0 s\u1ef1 t\u00e1c d\u1ee5ng c\u1ee7a oxi v\u1edbi m\u1ed9t ch\u1ea5t<br\/>$\\to$ ch\u1ea5t kh\u1eed hay c\u00f2n g\u1ecdi l\u00e0 ch\u1ea5t b\u1ecb oxi h\u00f3a \u0111\u1ec3 t\u1ea1o th\u00e0nh h\u1ee3p ch\u1ea5t v\u1edbi oxi<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2305},{"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 ph\u1ea3n \u1ee9ng sau: $2Al+F{{e}_{2}}{{O}_{3}}\\xrightarrow{{{t}^{o}}}2Fe+A{{l}_{2}}{{O}_{3}}$. Ch\u1ea5t kh\u1eed v\u00e0 ch\u1ea5t oxi h\u00f3a l\u00e0: ","select":["A. Ch\u1ea5t kh\u1eed l\u00e0 $Al$ v\u00e0 ch\u1ea5t oxi h\u00f3a l\u00e0 $F{{e}_{2}}{{O}_{3}}$ ","B. Ch\u1ea5t kh\u1eed l\u00e0 $F{{e}_{2}}{{O}_{3}}$ v\u00e0 ch\u1ea5t oxi h\u00f3a l\u00e0 $Al$ ","C. Ch\u1ea5t kh\u1eed l\u00e0 $Fe$ v\u00e0 ch\u1ea5t oxi h\u00f3a l\u00e0 $A{{l}_{2}}{{O}_{3}}$ ","D. Ch\u1ea5t kh\u1eed l\u00e0 $A{{l}_{2}}{{O}_{3}}$ v\u00e0 ch\u1ea5t oxi h\u00f3a l\u00e0 $Fe$ "],"hint":"","explain":"<span class='basic_left'>Ch\u1ea5t kh\u1eed l\u00e0 ch\u1ea5t chi\u1ebfm oxi c\u1ee7a ch\u1ea5t kh\u00e1c<br\/>Ch\u1ea5t oxi h\u00f3a l\u00e0 ch\u1ea5t nh\u01b0\u1eddng oxi cho ch\u1ea5t kh\u00e1c<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2306},{"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 h\u1ed7n h\u1ee3p $2,24$ l\u00edt ${{H}_{2}}$ v\u00e0 $7,392$ l\u00edt ${{O}_{2}}$ (\u0111ktc) ph\u1ea3n \u1ee9ng v\u1edbi nhau thu \u0111\u01b0\u1ee3c $m$ gam n\u01b0\u1edbc. Gi\u00e1 tr\u1ecb $m$ l\u00e0: ","select":["A. $3,6$ gam ","B. $3,9$ gam ","C. $1,8$ gam ","D. $6,6$ gam "],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{n}_{{{H}_{2}}}}=\\dfrac{2,24}{22,4}=0,1(mol) \\\\ & {{n}_{{{O}_{2}}}}=\\dfrac{7,392}{22,4}=0,33(mol) \\\\ & 2{{H}_{2}}+{{O}_{2}}\\xrightarrow{{{t}^{0}}}2{{H}_{2}}O \\\\ & \\to \\dfrac{0,1}{2}<\\dfrac{0,33}{1}(1) \\\\ \\end{aligned}$<br\/>T\u1eeb (1) suy ra sau ph\u1ea3n \u1ee9ng ${{O}_{2}}$ d\u01b0 v\u00e0 ${{H}_{2}}$ ph\u1ea3n \u1ee9ng h\u1ebft<br\/>$\\to$ l\u01b0\u1ee3ng n\u01b0\u1edbc sinh ra t\u00ednh theo l\u01b0\u1ee3ng h\u1ebft<br\/>Theo ph\u01b0\u01a1ng tr\u00ecnh:<br\/>$\\begin{aligned} & {{n}_{{{H}_{2}}O}}={{n}_{{{H}_{2}}}}=0,1(mol) \\\\ & \\to {{m}_{{{H}_{2}}O}}=0,1.18=1,8(g) \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2307},{"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 $48$ gam $CuO$ t\u00e1c d\u1ee5ng v\u1edbi ${{H}_{2}}$ khi \u0111un n\u00f3ng. Th\u1ec3 t\u00edch ${{H}_{2}}$ cho ph\u1ea3n \u1ee9ng tr\u00ean \u1edf \u0111ktc l\u00e0: ","select":["A. $22,4$ l\u00edt ","B. $44,8$ l\u00edt ","C. $3,36$ l\u00edt ","D. $13,44$ l\u00edt "],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{n}_{CuO}}=\\dfrac{48}{80}=0,6(mol) \\\\ & CuO+{{H}_{2}}\\xrightarrow{{{t}^{o}}}Cu+{{H}_{2}}O \\\\ \\end{aligned}$<br\/>Theo ph\u01b0\u01a1ng tr\u00ecnh h\u00f3a h\u1ecdc: <br\/>$\\begin{aligned} & {{n}_{{{H}_{2}}}}={{n}_{CuO}}=0,6(mol) \\\\ & \\to {{V}_{{{H}_{2}}}}=0,6.22,4=13,44(l) \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2308},{"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 kh\u00ed ${{H}_{2}}$ t\u00e1c d\u1ee5ng v\u1edbi $F{{e}_{2}}{{O}_{3}}$ \u0111un n\u00f3ng thu \u0111\u01b0\u1ee3c $11,2$ gam $Fe$. Kh\u1ed1i l\u01b0\u1ee3ng $F{{e}_{2}}{{O}_{3}}$ \u0111\u00e3 tham gia ph\u1ea3n \u1ee9ng l\u00e0: ","select":["A. $16$ gam ","B. $32$ gam ","C. $48$ gam ","D. $54$ gam "],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{n}_{Fe}}=\\dfrac{11,2}{56}=0,2(mol) \\\\ & F{{e}_{2}}{{O}_{3}}+3{{H}_{2}}\\xrightarrow{{{t}^{o}}}2Fe+3{{H}_{2}}O \\\\ \\end{aligned}$<br\/>Theo ph\u01b0\u01a1ng tr\u00ecnh h\u00f3a h\u1ecdc: <br\/>$\\begin{aligned} & {{n}_{F{{e}_{2}}{{O}_{3}}}}=\\dfrac{1}{2}{{n}_{Fe}}=0,1(mol) \\\\ & \\to {{m}_{F{{e}_{2}}{{O}_{3}}}}=0,1.160=16(g) \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2309},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"C\u00e1c ph\u1ea3n \u1ee9ng cho d\u01b0\u1edbi \u0111\u00e2y, ph\u1ea3n \u1ee9ng n\u00e0o l\u00e0 ph\u1ea3n \u1ee9ng oxi - h\u00f3a kh\u1eed? ","select":["A. $2Na+C{{l}_{2}}\\to 2NaCl$ ","B. $CaO+{{H}_{2}}O\\to Ca{{(OH)}_{2}}$ ","C. $HN{{O}_{3}}+KOH\\to KN{{O}_{3}}+{{H}_{2}}O$ ","D. $ZnO+2HCl\\to ZnC{{l}_{2}}+{{H}_{2}}O$ "],"hint":"","explain":"<span class='basic_left'>Ph\u1ea3n \u1ee9ng oxi h\u00f3a kh\u1eed l\u00e0 ph\u1ea3n \u1ee9ng h\u00f3a h\u1ecdc trong \u0111\u00f3 x\u1ea3y ra \u0111\u1ed3ng th\u1eddi s\u1ef1 oxi h\u00f3a v\u00e0 s\u1ef1 kh\u1eed.<br\/>Ngo\u00e0i ra ph\u1ea3n \u1ee9ng oxi h\u00f3a kh\u1eed c\u00f2n \u0111\u01b0\u1ee3c \u0111\u1ecbnh ngh\u0129a l\u00e0 ph\u1ea3n \u1ee9ng h\u00f3a h\u1ecdc trong \u0111\u00f3 c\u00f3 s\u1ef1 chuy\u1ec3n d\u1ecbch electron gi\u1eefa c\u00e1c ch\u1ea5t ph\u1ea3n \u1ee9ng<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2310},{"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 $13$ gam $Zn$ v\u00e0o dung d\u1ecbch ch\u1ee9a $0,5$ mol $HCl$. Sau ph\u1ea3n \u1ee9ng ch\u1ea5t n\u00e0o c\u00f2n d\u01b0 v\u00e0 d\u01b0 bao nhi\u00eau mol? ","select":["A. $Zn$ d\u01b0 v\u00e0 d\u01b0 $0,1$ mol ","B. $HCl$ d\u01b0 v\u00e0 d\u01b0 $0,3$ mol ","C. $HCl$ d\u01b0 v\u00e0 d\u01b0 $0,1$ mol ","D. $Zn$ d\u01b0 v\u00e0 d\u01b0 $0,2$ mol "],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{n}_{Zn}}=\\dfrac{13}{65}=0,2(mol) \\\\ & Zn+2HCl\\to ZnC{{l}_{2}}+{{H}_{2}} \\\\ \\end{aligned}$ <br\/>L\u1eadp t\u1ef7 l\u1ec7 s\u1ed1 mol c\u00e1c ch\u1ea5t tham gia ph\u1ea3n \u1ee9ng theo h\u1ec7 s\u1ed1 ph\u01b0\u01a1ng tr\u00ecnh ta th\u1ea5y: $\\dfrac{{{n}_{Zn}}}{1}<\\dfrac{{{n}_{HCl}}}{2}$ <br\/>V\u1eady sau ph\u1ea3n \u1ee9ng HCl d\u01b0, Zn ph\u1ea3n \u1ee9ng h\u1ebft<br\/>Theo ph\u01b0\u01a1ng tr\u00ecnh l\u01b0\u1ee3ng HCl tham gia ph\u1ea3n \u1ee9ng l\u00e0 ${{n}_{HCl}}=2{{n}_{Zn}}=0,4(mol)$ <br\/>$\\to$ Ph\u1ea7n mol HCl c\u00f2n l\u1ea1i sau ph\u1ea3n \u1ee9ng l\u00e0 ${{n}_{HCl(du)}}=0,5-0,4=0,1(mol)$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2311},{"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 $16,8$ gam b\u1ed9t s\u1eaft v\u00e0o dung d\u1ecbch ch\u1ee9a $0,6$ mol ${{H}_{2}}S{{O}_{4}}$. Th\u1ec3 t\u00edch kh\u00ed ${{H}_{2}}$ thu \u0111\u01b0\u1ee3c sau ph\u1ea3n \u1ee9ng \u1edf \u0111ktc l\u00e0: ","select":["A. $5,6$ l\u00edt ","B. $4,48$ l\u00edt ","C. $3,36$ l\u00edt ","D. $6,72$ l\u00edt "],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{n}_{Fe}}=\\dfrac{16,8}{56}=0,3(mol) \\\\ & Fe+{{H}_{2}}S{{O}_{4}}\\to FeS{{O}_{4}}+{{H}_{2}} \\\\ \\end{aligned}$<br\/>Theo ph\u01b0\u01a1ng tr\u00ecnh c\u1ee9 $1$ mol $Fe$ t\u00e1c d\u1ee5ng v\u1edbi $1$ mol ${{H}_{2}}S{{O}_{4}}$ t\u1ee9c l\u00e0 theo t\u1ec9 l\u1ec7 1: 1<br\/>M\u00e0 theo \u0111\u1ec1 b\u00e0i s\u1ed1 mol c\u1ee7a $Fe$ nh\u1ecf h\u01a1n s\u1ed1 mol c\u1ee7a ${{H}_{2}}S{{O}_{4}}$<br\/>$\\to$ sau ph\u1ea3n \u1ee9ng ${{H}_{2}}S{{O}_{4}}$ d\u01b0, Fe ph\u1ea3n \u1ee9ng h\u1ebft<br\/>L\u01b0\u1ee3ng kh\u00ed ${{H}_{2}}$ thu \u0111\u01b0\u1ee3c sau ph\u1ea3n \u1ee9ng \u0111\u01b0\u1ee3c t\u00ednh theo l\u01b0\u1ee3ng h\u1ebft l\u00e0 $Fe$<br\/>Theo PTHH:<br\/>$\\begin{aligned} & {{n}_{{{H}_{2}}}}={{n}_{Fe}}=0,3(mol) \\\\ & \\to {{V}_{{{H}_{2}}}}=0,3.22,4=6,72(l) \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2312},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"Kh\u1eed $5,8$ gam oxit s\u1eaft t\u1eeb $F{{e}_{3}}{{O}_{4}}$ b\u1eb1ng kh\u00ed hi\u0111ro cho $3,36 $ gam s\u1eaft. Hi\u1ec7u su\u1ea5t c\u1ee7a ph\u1ea3n \u1ee9ng l\u00e0: ","select":["A. $90\\%$ ","B. $80\\%$ ","C. $70\\%$ ","D. $85\\%$ "],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{n}_{F{{e}_{3}}{{O}_{4}}}}=\\dfrac{5,8}{232}=0,025(mol) \\\\ & {{n}_{Fe}}=\\dfrac{3,36}{56}=0,06(mol) \\\\ & F{{e}_{3}}{{O}_{4}}+4{{H}_{2}}\\xrightarrow{{{t}^{o}}}3Fe+4{{H}_{2}}O \\\\ \\end{aligned}$ <br\/>\u0110\u1ea7u ti\u00ean ta t\u00ednh l\u01b0\u1ee3ng tham gia ph\u1ea3n \u1ee9ng c\u1ee7a $F{{e}_{3}}{{O}_{4}}$ theo s\u1ea3n ph\u1ea9m l\u00e0 Fe: ${{n}_{F{{e}_{3}}{{O}_{4}}}}=\\dfrac{1}{3}{{n}_{Fe}}=0,02(mol)$<br\/>$\\to$ L\u01b0\u1ee3ng $F{{e}_{3}}{{O}_{4}}$ tham gia ph\u1ea3n \u1ee9ng nh\u1ecf h\u01a1n so v\u1edbi ban \u0111\u1ea7u, \u0111i\u1ec1u n\u00e0y ch\u00fang t\u1ecf ph\u1ea3n \u1ee9ng x\u1ea3y ra kh\u00f4ng ho\u00e0n to\u00e0n<br\/>V\u1eady hi\u1ec7u su\u1ea5t c\u1ee7a ph\u1ea3n \u1ee9ng l\u00e0 <br\/>$Hs=\\dfrac{0,02}{0,025}.100\\%=80\\%$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2313},{"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 ph\u00f2ng th\u00ed nghi\u1ec7m, ng\u01b0\u1eddi ta \u0111i\u1ec1u ch\u1ebf ${{H}_{2}}$ b\u1eb1ng c\u00e1ch cho $1,08$ gam m\u1ed9t kim lo\u1ea1i $R$ h\u00f3a tr\u1ecb III t\u00e1c d\u1ee5ng v\u1edbi axit $HCl$ thu \u0111\u01b0\u1ee3c $1,344$ l\u00edt kh\u00ed hi\u0111ro \u1edf \u0111ktc. Kim lo\u1ea1i $R$ l\u00e0: ","select":["A. Fe ","B. Cr ","C. Al ","D. Cu "],"hint":"","explain":"<span class='basic_left'>Ph\u01b0\u01a1ng tr\u00ecnh h\u00f3a h\u1ecdc t\u1ed5ng qu\u00e1t d\u1ea1ng: <br\/>$2R+6HCl\\to 2RC{{l}_{3}}+3{{H}_{2}}$<br\/>${{n}_{{{H}_{2}}}}=\\dfrac{1,344}{22,4}=0,06(mol)$<br\/>Theo ph\u01b0\u01a1ng tr\u00ecnh h\u00f3a h\u1ecdc: ${{n}_{R}}=\\dfrac{2}{3}{{n}_{{{H}_{2}}}}=0,04(mol)$<br\/>$\\to$ Kh\u1ed1i l\u01b0\u1ee3ng mol c\u1ee7a R l\u00e0 ${{M}_{R}}=\\dfrac{1,08}{0,04}=27$<br\/>V\u1eady $R$ l\u00e0 nh\u00f4m (Al)<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2314},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"Ph\u1ea3n \u1ee9ng n\u00e0o d\u01b0\u1edbi \u0111\u00e2y kh\u00f4ng ph\u1ea3i l\u00e0 ph\u1ea3n \u1ee9ng th\u1ebf? ","select":["A. $CuO+{{H}_{2}}\\xrightarrow{{{t}^{o}}}Cu+{{H}_{2}}O$ ","B. $Mg+2HCl\\to MgC{{l}_{2}}+{{H}_{2}}$ ","C. $Zn+CuS{{O}_{4}}\\to ZnS{{O}_{4}}+Cu$ ","D. $Ca{{(OH)}_{2}}+C{{O}_{2}}\\to CaC{{O}_{3}}+{{H}_{2}}O$ "],"hint":"","explain":"<span class='basic_left'>Ph\u1ea3n \u1ee9ng th\u1ebf l\u00e0 ph\u1ea3n \u1ee9ng h\u00f3a h\u1ecdc gi\u1eefa \u0111\u01a1n ch\u1ea5t v\u00e0 h\u1ee3p ch\u1ea5t, trong \u0111\u00f3 nguy\u00ean t\u1eed c\u1ee7a \u0111\u01a1n ch\u1ea5t thay th\u1ebf nguy\u00ean t\u1eed c\u1ee7a m\u1ed9t nguy\u00ean t\u1ed1 kh\u00e1c trong h\u1ee3p ch\u1ea5t<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2315},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"Ng\u01b0\u1eddi ta d\u00f9ng kh\u00ed cacbon oxit \u0111\u1ec3 kh\u1eed \u0111\u1ed3ng(II) oxit. N\u1ebfu kh\u1eed $150$ gam \u0111\u1ed3ng(II) oxit th\u00ec th\u1ec3 t\u00edch kh\u00ed CO c\u1ea7n d\u00f9ng \u1edf \u0111ktc l\u00e0: ","select":["A. $50$ l\u00edt ","B. $46$ l\u00edt ","C. $42$ l\u00edt ","D. $36$ l\u00edt "],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{n}_{CuO}}=\\dfrac{150}{80}=1,875(mol) \\\\ & CuO+CO\\xrightarrow{{{t}^{0}}}Cu+C{{O}_{2}} \\\\ \\end{aligned}$ <br\/>Theo ph\u01b0\u01a1ng tr\u00ecnh ph\u1ea3n \u1ee9ng:<br\/>$\\begin{aligned} & {{n}_{CO}}={{n}_{CuO}}=1,875(mol) \\\\ & \\to {{V}_{CO}}=1,875.22,4=42(l) \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2316},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"M\u1ed9t h\u1ed7n h\u1ee3p g\u1ed3m $60\\%F{{e}_{2}}{{O}_{3}}$ v\u00e0 $40\\%CuO$ v\u1ec1 kh\u1ed1i l\u01b0\u1ee3ng. Ng\u01b0\u1eddi ta d\u00f9ng ${{H}_{2}}$ \u0111\u1ec3 kh\u1eed $20$ gam h\u1ed7n h\u1ee3p tr\u00ean. S\u1ed1 mol ${{H}_{2}}$ \u0111\u00e3 tham gia ph\u1ea3n \u1ee9ng l\u00e0: ","select":["A. $0,425$ mol ","B. $0,325$ mol ","C. $0,456$ mol ","D. $0,784$ mol "],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{m}_{F{{e}_{2}}{{O}_{3}}}}=\\dfrac{20.60}{100}=12(g)\\to {{n}_{F{{e}_{2}}{{O}_{3}}}}=\\dfrac{12}{160}=0,075(mol) \\\\ & {{m}_{CuO}}=20-12=8(g)\\to {{n}_{CuO}}=\\dfrac{8}{80}=0,1(mol) \\\\ & F{{e}_{2}}{{O}_{3}}+3{{H}_{2}}\\xrightarrow{{{t}^{o}}}2Fe+3{{H}_{2}}O(1) \\\\ & CuO+{{H}_{2}}\\xrightarrow{{{t}^{o}}}CuO+{{H}_{2}}O(2) \\\\ \\end{aligned}$<br\/>Theo ph\u01b0\u01a1ng tr\u00ecnh (1) v\u00e0 (2) ta c\u00f3:<br\/>${{n}_{{{H}_{2}}}}={{n}_{{{H}_{2}}(1)}}+{{n}_{{{H}_{2}}(2)}}=3{{n}_{F{{e}_{2}}{{O}_{3}}}}+{{n}_{CuO}}=3.0,075+0,1=0,325(mol)$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2317},{"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 m\u1ea1t s\u1eaft v\u00e0o m\u1ed9t dung d\u1ecbch ch\u1ee9a $600$ ml ${{H}_{2}}S{{O}_{4}}$ lo\u00e3ng, d\u01b0. Sau m\u1ed9t th\u1eddi gian, b\u1ed9t s\u1eaft tan ho\u00e0n to\u00e0n v\u00e0 ng\u01b0\u1eddi ta thu \u0111\u01b0\u1ee3c $1,232$ l\u00edt kh\u00ed hi\u0111ro \u1edf \u0111ktc. \u0110\u1ec3 c\u00f3 l\u01b0\u1ee3ng s\u1eaft tham gia ph\u1ea3n \u1ee9ng tr\u00ean, ph\u1ea3i cho bao nhi\u00eau gam s\u1eaft(III) oxit t\u00e1c d\u1ee5ng v\u1edbi kh\u00ed hi\u0111ro ? ","select":["A. $4,4$ gam ","B. $3,3$ gam ","C. $4,6$ gam ","D. $6$ gam "],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{n}_{{{H}_{2}}}}=\\dfrac{1,232}{22,4}=0,055(mol) \\\\ & Fe+{{H}_{2}}S{{O}_{4}}\\to FeS{{O}_{4}}+{{H}_{2}} \\\\ \\end{aligned}$<br\/>Theo ph\u01b0\u01a1ng tr\u00ecnh c\u1ee9 1 mol Fe tham gia ph\u1ea3n \u1ee9ng sinh ra 1 mol kh\u00ed hi\u0111ro<br\/>$\\to {{n}_{Fe}}={{n}_{{{H}_{2}}}}=0,055(mol)$<br\/>$\\begin{aligned} & F{{e}_{2}}{{O}_{3}}+3{{H}_{2}}\\xrightarrow{{{t}^{o}}}2Fe+3{{H}_{2}}O \\\\ & \\begin{matrix} 0,0275mol & \\leftarrow & 0,055mol & {} & {} & {} & {} \\\\\\end{matrix} \\\\ & \\to {{m}_{F{{e}_{2}}{{O}_{3}}}}=0,0275.160=4,4(g) \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2318},{"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 $5,04$ gam s\u1eaft t\u00e1c d\u1ee5ng v\u1edbi dung d\u1ecbch lo\u00e3ng c\u00f3 ch\u1ee9a $6,86$ gam axit sunfuric. Sau ph\u1ea3n \u1ee9ng ch\u1ea5t n\u00e0o c\u00f2n d\u01b0 v\u00e0 d\u01b0 bao nhi\u00eau gam? ","select":["A. $Fe$ d\u01b0 v\u00e0 d\u01b0 $11,2$ gam ","B. ${{H}_{2}}S{{O}_{4}}$ d\u01b0 v\u00e0 d\u01b0 $1,96$ gam ","C. ${{H}_{2}}S{{O}_{4}}$ d\u01b0 v\u00e0 d\u01b0 $19,6$ gam ","D. $Fe$ d\u01b0 v\u00e0 d\u01b0 $1,12$ gam "],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{n}_{Fe}}=\\dfrac{5,04}{56}=0,09(mol) \\\\ & {{n}_{{{H}_{2}}S{{O}_{4}}}}=\\frac{6,86}{98}=0,07(mol) \\\\ & Fe+{{H}_{2}}S{{O}_{4}}\\to FeS{{O}_{4}}+{{H}_{2}} \\\\ \\end{aligned}$<br\/>Theo ph\u01b0\u01a1ng tr\u00ecnh ph\u1ea3n \u1ee9ng c\u1ee9 1 mol Fe t\u00e1c d\u1ee5ng v\u1edbi 1 mol axit sunfuric<br\/>Theo t\u1ec9 l\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh s\u1ed1 mol Fe l\u1edbn h\u01a1n s\u1ed1 mol axit sunfuric<br\/>$\\to$ Fe d\u01b0, ${{H}_{2}}S{{O}_{4}}$ ph\u1ea3n \u1ee9ng h\u1ebft <br\/>L\u01b0\u1ee3ng Fe d\u01b0 l\u00e0 0,09 - 0,07 = 0,02 mol<br\/>$\\to$ Kh\u1ed1i l\u01b0\u1ee3ng Fe c\u00f2n l\u1ea1i sau ph\u1ea3n \u1ee9ng l\u00e0 ${{m}_{Fe}}=0,02.56=1,12(g)$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2319}],"lesson":{"save":0,"level":2}}