Bài tập trung bình bài Bài 19: Sắt

Home Lớp 9 Hóa học 9 Bài 19: Sắt Bài tập luyện

{"segment":[{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Ho\u00e0 tan h\u1ed7n h\u1ee3p g\u1ed3m Fe v\u00e0 Cu v\u00e0o dung d\u1ecbch HCl (v\u1eeba \u0111\u1ee7). C\u00e1c s\u1ea3n ph\u1ea9m thu \u0111\u01b0\u1ee3c sau ph\u1ea3n \u1ee9ng l\u00e0","select":["A. $FeCl_2$ v\u00e0 kh\u00ed $H_2$","B. $FeCl_2$, $Cu$ v\u00e0 kh\u00ed $H_2$","C. $Cu$ v\u00e0 kh\u00ed $H_2$ ","D. $FeCl_2$ v\u00e0 $Cu$"],"hint":"","explain":"$Fe+2HCl\\to FeC{{l}_{2}}+{{H}_{2}}\\uparrow $ <br\/>Cu kh\u00f4ng t\u00e1c d\u1ee5ng v\u1edbi dung d\u1ecbch HCl<br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2071},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Cho d\u00e2y s\u1eaft qu\u1ea5n h\u00ecnh l\u00f2 xo (\u0111\u00e3 \u0111\u01b0\u1ee3c nung n\u00f3ng \u0111\u1ecf) v\u00e0o l\u1ecd \u0111\u1ef1ng kh\u00ed clo. Hi\u1ec7n t\u01b0\u1ee3ng x\u1ea3y ra l\u00e0","select":["A. S\u1eaft ch\u00e1y t\u1ea1o th\u00e0nh kh\u00f3i tr\u1eafng d\u00e0y \u0111\u1eb7t b\u00e1m v\u00e0o th\u00e0nh b\u00ecnh.A. m\u00e9t ","B. Kh\u00f4ng th\u1ea5y hi\u1ec7n t\u01b0\u1ee3ng ph\u1ea3n \u1ee9ng","C. S\u1eaft ch\u00e1y s\u00e1ng t\u1ea1o th\u00e0nh kh\u00f3i m\u00e0u n\u00e2u \u0111\u1ecf ","D. S\u1eaft ch\u00e1y s\u00e1ng t\u1ea1o th\u00e0nh kh\u00f3i m\u00e0u \u0111en"],"hint":"","explain":"S\u1eaft ch\u00e1y s\u00e1ng t\u1ea1o th\u00e0nh kh\u00f3i m\u00e0u n\u00e2u \u0111\u1ecf do t\u1ea1o th\u00e0nh $FeCl_3$ <br\/>$2Fe+3C{{l}_{2}}\\xrightarrow{{{t}^{0}}}2FeC{{l}_{3}}$ <br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":1}]}],"id_ques":2072},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" S\u1eaft t\u00e1c d\u1ee5ng v\u1edbi oxi \u1edf nhi\u1ec7t \u0111\u1ed9 cao t\u1ea1o th\u00e0nh ","select":["A. S\u1eaft (II) oxit ","B. S\u1eaft (III) oxit","C. S\u1eaft t\u1eeb oxit ","D. Kh\u00f4ng ph\u1ea3n \u1ee9ng"],"hint":"","explain":"$3Fe+2{{O}_{2}}\\xrightarrow{{{t}^{0}}}F{{e}_{3}}{{O}_{4}}$ <br\/>$Fe_3O_4$: s\u1eaft t\u1eeb oxit.<br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2073},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" Ng\u00e2m m\u1ed9t c\u00e2y \u0111inh s\u1eaft s\u1ea1ch v\u00e0o dung d\u1ecbch b\u1ea1c nitrat. Hi\u1ec7n t\u01b0\u1ee3ng x\u1ea3y ra l\u00e0:","select":["A. Kh\u00f4ng c\u00f3 hi\u1ec7n t\u01b0\u1ee3ng g\u00ec c\u1ea3. ","B. B\u1ea1c \u0111\u01b0\u1ee3c gi\u1ea3i ph\u00f3ng, nh\u01b0ng s\u1eaft kh\u00f4ng bi\u1ebfn \u0111\u1ed5i.","C. Kh\u00f4ng c\u00f3 ch\u1ea5t n\u00e0o sinh ra, ch\u1ec9 c\u00f3 s\u1eaft b\u1ecb ho\u00e0 tan.","D. S\u1eaft b\u1ecb ho\u00e0 tan m\u1ed9t ph\u1ea7n, b\u1ea1c \u0111\u01b0\u1ee3c gi\u1ea3i ph\u00f3ng."],"hint":"","explain":"S\u1eaft tan d\u1ea7n do t\u00e1c d\u1ee5ng v\u1edbi dung d\u1ecbch b\u1ea1c nitrat t\u1ea1o th\u00e0nh b\u1ea1c <br\/>$2AgN{{O}_{3}}+Fe\\to Fe{{(N{{O}_{3}})}_{2}}+2Ag$ <br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":1}]}],"id_ques":2074},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Cho 1 \u0111inh s\u1eaft v\u00e0o dung d\u1ecbch $HNO_3$ \u0111\u1eb7c ngu\u1ed9i. Hi\u1ec7n t\u01b0\u1ee3ng quan s\u00e1t \u0111\u01b0\u1ee3c l\u00e0:","select":["A. C\u00f3 kh\u00ed n\u00e2u \u0111\u1ecf bay ra","B. C\u00f3 kh\u00ed kh\u00f4ng m\u00e0u bay ra","C. Kh\u00f4ng c\u00f3 hi\u1ec7n t\u01b0\u1ee3ng g\u00ec","D. Dung d\u1ecbch m\u00e0u xanh lam, c\u00f3 kh\u00ed m\u00e0u n\u00e2u \u0111\u1ecf."],"hint":"","explain":"S\u1eaft kh\u00f4ng t\u00e1c d\u1ee5ng v\u1edbi $H_SO_4$ \u0111\u1eb7c ngu\u1ed9i, $HNO_3$ \u0111\u1eb7c ngu\u1ed9i.<br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":1}]}],"id_ques":2075},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" S\u1eaft ho\u1ea1t \u0111\u1ed9ng h\u00f3a h\u1ecdc m\u1ea1nh h\u01a1n kim lo\u1ea1i n\u00e0o sau \u0111\u00e2y:","select":["A. Mg","B. Al","C. Cu ","D. Zn"],"hint":"","explain":"D\u00e3y ho\u1ea1t \u0111\u1ed9ng h\u00f3a h\u1ecdc c\u1ee7a kim lo\u1ea1i: <br\/>K, Na, Mg, Al, Zn, Fe, Pb, H, Cu, Ag, Au<br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2076},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" S\u1eaft t\u00e1c d\u1ee5ng v\u1edbi l\u01b0u hu\u1ef3nh \u1edf nhi\u1ec7t \u0111\u1ed9 cao s\u1ea3n ph\u1ea9m thu \u0111\u01b0\u1ee3c l\u00e0:","select":["A. $FeS$","B. $Fe_2S_3$","C. $FeS_2$ ","D. $Fe_2S$"],"hint":"","explain":"$Fe+S\\xrightarrow{{{t}^{0}}}FeS$ <br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2077},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Mu\u1ed1n chuy\u1ec3n $Fe$ th\u00e0nh $FeCl_3$ ta cho $Fe$ t\u00e1c d\u1ee5ng v\u1edbi","select":["A. $HCl$ ","B. $CuCl_2$","C. $Cl_2$ ","D. $AlCl_3$"],"hint":"","explain":"$2Fe+3C{{l}_{2}}\\xrightarrow{{{t}^{0}}}2FeC{{l}_{3}}$<br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2078},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" S\u1eaft l\u00e0 kim lo\u1ea1i","select":["A. M\u00e0u tr\u1eafng xanh, d\u1eabn \u0111i\u1ec7n t\u1ed1t h\u01a1n nh\u00f4m","B. M\u00e0u tr\u1eafng x\u00e1m, d\u1eabn \u0111i\u1ec7n t\u1ed1t h\u01a1n nh\u00f4m","C. M\u00e0u tr\u1eafng x\u00e1m, d\u1eabn \u0111i\u1ec7n k\u00e9m h\u01a1n nh\u00f4m","D. M\u00e0u tr\u1eafng xanh, d\u1eabn \u0111i\u1ec7n k\u00e9m h\u01a1n nh\u00f4m."],"hint":"","explain":"S\u1eaft l\u00e0 kim lo\u1ea1i m\u00e0u tr\u1eafng x\u00e1m, d\u1ebbo, d\u1eabn \u0111i\u1ec7n d\u1eabn nhi\u1ec7t t\u1ed1t. C\u00f3 t\u00ednh nhi\u1ec5m t\u1eeb. T\u00ednh d\u1eabn \u0111i\u1ec7n c\u1ee7a s\u1eaft k\u00e9m h\u01a1n nh\u00f4m.<br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":1}]}],"id_ques":2079},{"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":"$\\begin{aligned}& {{n}_{F{{e}_{2}}{{O}_{3}}}}=0,075\\,mol \\\\ & F{{e}_{2}}{{O}_{3}}+3{{H}_{2}}\\xrightarrow{{{t}^{0}}}2Fe+3{{H}_{2}}O \\\\ & {{n}_{{{H}_{2}}}}=3{{n}_{F{{e}_{2}}{{O}_{3}}}}=0,225\\,mol \\\\ & {{V}_{{{H}_{2}}}}=5,04\\,l \\\\ \\end{aligned}$ <br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2080},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" H\u00f3a tr\u1ecb c\u1ee7a Fe l\u00e0","select":["A. II","B. III ","C. II ho\u1eb7c III ","D. III ho\u1eb7c IV"],"hint":"","explain":"Fe c\u00f3 2 h\u00f3a tr\u1ecb l\u00e0 II, III<br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2081},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" \u0110\u1ec3 chuy\u1ec3n $FeCl_3$ th\u00e0nh $Fe(OH)_3$, ng\u01b0\u1eddi ta d\u00f9ng dung d\u1ecbch","select":["A. $HCl$ ","B. $H_2SO_4$","C. $NaOH$","D. $AgNO_3$"],"hint":"","explain":"$FeC{{l}_{3}}+3NaOH\\to 3NaCl+Fe{{(OH)}_{3}}\\downarrow $ <br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2082},{"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 ch\u1ea5t sau \u0111\u00e2y ch\u1ea5t n\u00e0o ch\u1ee9a h\u00e0m l\u01b0\u1ee3ng s\u1eaft nhi\u1ec1u nh\u1ea5t","select":["A. $FeS_2$ ","B. $FeO$","C. $Fe_2O_3$ ","D. $Fe_3O_4$"],"hint":"","explain":"$\\begin{aligned}& A.\\%{{m}_{Fe}}=\\dfrac{56}{120}.100\\%\\approx 47\\% \\\\ & B.\\%{{m}_{Fe}}=\\dfrac{56}{72}.100\\%\\approx 78\\% \\\\ & C.\\%{{m}_{Fe}}=\\dfrac{56.2}{160}.100\\%=70\\% \\\\ & D.\\%{{m}_{Fe}}=\\dfrac{56.3}{232}.100\\%\\approx 72\\% \\\\ \\end{aligned}$ <br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2083},{"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 t\u1ea1o ra mu\u1ed1i s\u1eaft (III) sunfat l\u00e0:","select":["A. S\u1eaft ph\u1ea3n \u1ee9ng v\u1edbi $H_2SO_4$ \u0111\u1eb7c, n\u00f3ng.","B. S\u1eaft ph\u1ea3n \u1ee9ng v\u1edbi dung d\u1ecbch $H_2SO_4$ lo\u00e3ng ","C. S\u1eaft ph\u1ea3n \u1ee9ng v\u1edbi dung d\u1ecbch $CuSO_4$ ","D. S\u1eaft ph\u1ea3n \u1ee9ng v\u1edbi dung d\u1ecbch $Al_2(SO_4)_3$"],"hint":"","explain":"$2Fe+6{{H}_{2}}S{{O}_{4\\text{\u0111\u1eb7c}}}\\xrightarrow{{{t}^{0}}}F{{e}_{2}}{{(S{{O}_{4}})}_{3}}+3S{{O}_{2}}+6{{H}_{2}}O$ <br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":1}]}],"id_ques":2084},{"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 m\u1ed9t l\u01b0\u1ee3ng qu\u1eb7ng hemantit ch\u1ee9a 80% ($Fe_2O_3$) thu \u0111\u01b0\u1ee3c 1,68 t\u1ea5n s\u1eaft, kh\u1ed1i l\u01b0\u1ee3ng qu\u1eb7ng c\u1ea7n l\u1ea5y l\u00e0","select":["A. 2,4 t\u1ea5n","B. 2,6 t\u1ea5n","C. 2,8 t\u1ea5n ","D. 3,0 t\u1ea5n"],"hint":"","explain":"$\\begin{aligned} & F{{e}_{2}}{{O}_{3}}+3{{H}_{2}}\\xrightarrow{{{t}^{0}}}2Fe+3{{H}_{2}}O \\\\ & 160\\,\\text{t\u1ea5n}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,112\\,\\text{t\u1ea5n}\\, \\\\ & x\\,\\text{ t\u1ea5n }\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,1,68\\text{ t\u1ea5n }\\, \\\\ & x=\\dfrac{1,68.160}{112}=2,4\\text{ t\u1ea5n } \\\\ & {{m}_{\\text{qu\u1eb7ng}}}=\\dfrac{2,4.100\\%}{80\\%}=3\\text{ t\u1ea5n } \\\\ \\end{aligned}$<br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2085},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" \u0110\u1ed1t ch\u00e1y 1 mol s\u1eaft trong oxi, thu \u0111\u01b0\u1ee3c 1 mol s\u1eaft oxit. C\u00f4ng th\u1ee9c c\u1ee7a s\u1eaft oxit n\u00e0y l\u00e0:","select":["A. $FeO$ ","B. $Fe_2O_3$ ","C. $Fe_3O_4$","D. $Fe_4O_3$"],"hint":"","explain":"Ch\u1ec9 c\u00f3 \u0111\u00e1p \u00e1n A th\u1ecfa m\u00e3n: $\\dfrac{{{n}_{Fe}}}{{{n}_{Oxit}}}=\\dfrac{1}{1}$ <br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2086},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Cho 5,6g s\u1eaft tan ho\u00e0n to\u00e0n trong dd c\u00f3 ch\u1ee9a 0,2mol $H_2SO_4$. Th\u1ec3 t\u00edch $H_2$ thu \u0111c l\u00e0:","select":["A. 2,24 l\u00edt ","B. 5,04 l\u00edt.","C. 3 l\u00edt ","D. 7,72 l\u00edt"],"hint":"","explain":"$\\begin{aligned}& Fe+{{H}_{2}}S{{O}_{4}}\\to FeS{{O}_{4}}+{{H}_{2}}\\uparrow \\\\ & {{n}_{Fe}}=0,1\\,mol \\\\ & \\dfrac{{{n}_{Fe}}}{1}<\\dfrac{{{n}_{{{H}_{2}}S{{O}_{4}}}}}{1} \\\\ \\end{aligned}$ <br\/>=> $Fe$ h\u1ebft, t\u00ednh theo $Fe$ <br\/>$\\begin{aligned}& {{n}_{{{H}_{2}}}}={{n}_{Fe}}=0,1\\,mol \\\\ & {{V}_{{{H}_{2}}}}=2,24\\,l \\\\ \\end{aligned}$ <br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2087},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" M\u1ed9t lo\u1ea1i qu\u1eb7ng ch\u1ee9a 82% $Fe_2O_3$. Th\u00e0nh ph\u1ea7n ph\u1ea7n tr\u0103m c\u1ee7a $Fe$ trong qu\u1eb7ng theo kh\u1ed1i l\u01b0\u1ee3ng l\u00e0","select":["A. 54,7%","B. 57,0 %","C. 57,4% ","D. 56,4 %"],"hint":"","explain":"Gi\u1ea3 s\u1eed c\u00f3 100 gam qu\u1eb7ng <br\/>$\\begin{aligned}& =>{{m}_{F{{e}_{2}}{{O}_{3}}}}=\\dfrac{100.82\\%}{100\\%}=82\\,gam \\\\ & {{n}_{F{{e}_{2}}{{O}_{3}}}}=0,5125\\,mol \\\\ & {{n}_{Fe}}=2.{{n}_{F{{e}_{2}}{{O}_{3}}}}=1,025\\,mol \\\\ & {{m}_{Fe}}=57,4\\,gam \\\\ & \\%{{m}_{Fe}}=\\dfrac{57,4}{100}.100\\%=57,4\\% \\\\ \\end{aligned}$<br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2088},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Ng\u00e2m m\u1ed9t l\u00e1 s\u1eaft c\u00f3 kh\u1ed1i l\u01b0\u1ee3ng 20g v\u00e0o dung d\u1ecbch b\u1ea1c nitrat, sau m\u1ed9t th\u1eddi gian ph\u1ea3n \u1ee9ng nh\u1ea5c l\u00e1 kim lo\u1ea1i ra l\u00e0m kh\u00f4 c\u00e2n n\u1eb7ng 23,2g. L\u00e1 kim lo\u1ea1i sau ph\u1ea3n \u1ee9ng c\u00f3:","select":["A. 18,88g Fe v\u00e0 4,32g Ag ","B. 1,880g Fe v\u00e0 4,32g Ag","C. 15,68g Fe v\u00e0 4,32g Ag ","D. 18,88g Fe v\u00e0 3,42g Ag"],"hint":"","explain":"Kh\u1ed1i l\u01b0\u1ee3ng \u0111inh s\u1eaft t\u0103ng 1 l\u01b0\u01a1ng = 3,2 gam <br\/>G\u1ecdi x l\u00e0 s\u1ed1 mol c\u1ee7a Fe ph\u1ea3n \u1ee9ng <br\/>$\\begin{aligned}& Fe+2AgN{{O}_{3}}\\to Fe{{(N{{O}_{3}})}_{2}}+2Ag \\\\ & x\\,mol\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,2x\\,mol \\\\ \\end{aligned}$ <br\/>Kh\u1ed1i l\u01b0\u1ee3ng \u0111inh s\u1eaft t\u0103ng: <br\/>$\\begin{aligned}& ={{m}_{Ag}}-{{m}_{Fe}} \\\\ & =108.2x-56x \\\\ & =160x=3,2\\,gam \\\\ & =>x=0,02\\,mol \\\\ & =>{{m}_{Ag}}=0,02.2.108=4,32\\,gam \\\\ & =>{{m}_{Fe}}=18,88\\,gam \\\\ \\end{aligned}$ <br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":1}]}],"id_ques":2089},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Cho 3,25 gam s\u1eaft clorua (ch\u01b0a bi\u1ebft h\u00f3a tr\u1ecb) t\u00e1c d\u1ee5ng v\u1edbi dung d\u1ecbch $AgNO_3$ d\u01b0, thu \u0111\u01b0\u1ee3c 8,61 gam k\u1ebft t\u1ee7a. C\u00f4ng th\u1ee9c c\u1ee7a mu\u1ed1i s\u1eaft clorua n\u00e0o sau \u0111\u00e2y l\u00e0 \u0111\u00fang","select":["A. $FeCl_2$","B. $FeCl_3$","C. $FeCl_4$ ","D. $Fe_2Cl_3$"],"hint":"","explain":"G\u1ecdi c\u00f4ng th\u1ee9c c\u1ee7a s\u1eaft clorua l\u00e0 $FeCl_x$ <br\/>$\\begin{aligned}& FeC{{l}_{x}}+xAgN{{O}_{3}}\\to xAgCl\\downarrow +Fe{{(N{{O}_{3}})}_{x}} \\\\ & {{n}_{AgCl}}=0,06\\,mol \\\\ & {{n}_{FeC{{l}_{x}}}}=\\dfrac{1}{x}{{n}_{AgCl}}=\\dfrac{0,06}{x}\\,mol \\\\ & {{M}_{FeC{{l}_{x}}}}=\\dfrac{3,25}{\\dfrac{0,06}{x}}=\\dfrac{325}{6}x \\\\ & =>\\dfrac{325}{6}x=56+35,5x \\\\ & =>x=3 \\\\ \\end{aligned}$ <br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2090}],"lesson":{"save":0,"level":2}}

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Câu hỏi này theo dạng chọn đáp án đúng, sau khi đọc xong câu hỏi, bạn bấm vào một trong số các đáp án mà chương trình đưa ra bên dưới, sau đó bấm vào nút gửi để kiểm tra đáp án và sẵn sàng chuyển sang câu hỏi kế tiếp

Trả lời đúng trong khoảng thời gian quy định bạn sẽ được + số điểm như sau:
Trong khoảng 1 phút đầu tiên	+ 5 điểm
Trong khoảng 1 phút -> 2 phút	+ 4 điểm
Trong khoảng 2 phút -> 3 phút	+ 3 điểm
Trong khoảng 3 phút -> 4 phút	+ 2 điểm
Trong khoảng 4 phút -> 5 phút	+ 1 điểm

Quá 5 phút: không được cộng điểm

Tổng thời gian làm mỗi câu (không giới hạn)

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Bấm vào đây nếu phát hiện có lỗi hoặc muốn gửi góp ý

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