{"segment":[{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"\u0110\u1ed1t ch\u00e1y ho\u00e0n $5,2$ g $Zn$ trong oxi thu \u0111\u01b0\u1ee3c $ZnO$. Kh\u1ed1i l\u01b0\u1ee3ng $ZnO$ thu \u0111\u01b0\u1ee3c l\u00e0 ","select":["A. $8,48$ g ","B. $6,48$ g ","C. $7,48$ g ","D. $9,48$ g "],"hint":"","explain":"<span class='basic_left'>Ph\u01b0\u01a1ng tr\u00ecnh h\u00f3a h\u1ecdc c\u1ee7a ph\u1ea3n \u1ee9ng l\u00e0<br\/>$2Zn+{{O}_{2}}\\to 2ZnO$<br\/>S\u1ed1 mol c\u1ee7a $Zn$ l\u00e0<br\/>$n=\\dfrac{m}{M}=\\dfrac{5,2}{65}=0,08(mol)$<br\/>Theo ph\u01b0\u01a1ng tr\u00ecnh h\u00f3a h\u1ecdc ta c\u00f3:<br\/>C\u1ee9 2 mol $Zn$ tham gia ph\u1ea3n \u1ee9ng, s\u1ebd thu \u0111\u01b0\u1ee3c 2 mol $ZnO$<br\/>V\u1eady 0,08 mol $Zn$ tham gia ph\u1ea3n \u1ee9ng s\u1ebd thu \u0111\u01b0\u1ee3c x mol $ZnO$<br\/>$\\Rightarrow x=\\dfrac{0,08.2}{2}=0,08(mol)$<br\/>$\\Rightarrow {{m}_{ZnO}}=0,08.(65+16)=6,48(g)$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2110},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"\u0110\u1ec3 \u0111\u1ed1t ch\u00e1y ho\u00e0n to\u00e0n $a$ (g) $Al$ c\u1ea7n d\u00f9ng h\u1ebft $28,8$ g oxi. Ph\u1ea3n \u1ee9ng thu \u0111\u01b0\u1ee3c $x$ (g) $A{{l}_{2}}{{O}_{3}}$. Ph\u01b0\u01a1ng tr\u00ecnh c\u1ee7a ph\u1ea3n \u1ee9ng \u0111\u1ed1t ch\u00e1y l\u00e0 ","select":["A. $Al+3{{O}_{2}}\\to 2A{{l}_{2}}{{O}_{3}}$ ","B. $4Al+{{O}_{2}}\\to 2A{{l}_{2}}{{O}_{3}}$ ","C. $4Al+3{{O}_{2}}\\to 2A{{l}_{2}}{{O}_{3}}$ ","D. $4Al+3{{O}_{2}}\\to A{{l}_{2}}{{O}_{3}}$ "],"hint":"","explain":"<span class='basic_left'>S\u01a1 \u0111\u1ed3 c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh d\u1ea1ng: $Al+{{O}_{2}}--->A{{l}_{2}}{{O}_{3}}$<br\/>S\u1ed1 nguy\u00ean t\u1eed $Al$ v\u00e0 $O$ \u0111\u1ec1u kh\u00f4ng b\u1eb1ng nhau, nh\u01b0ng s\u1ed1 nguy\u00ean t\u1eed $O$ c\u00f3 s\u1ed1 nguy\u00ean t\u1eed nhi\u1ec1u h\u01a1n. Ta b\u1eaft \u0111\u1ea7u c\u00e2n b\u1eb1ng s\u1ed1 nguy\u00ean t\u1eed n\u00e0y<br\/>Tr\u01b0\u1edbc h\u1ebft l\u00e0m ch\u1eb5n s\u1ed1 nguy\u00ean t\u1eed $O$ \u1edf b\u00ean ph\u1ea3i, t\u1ee9c \u0111\u1eb7t h\u1ec7 s\u1ed1 2 tr\u01b0\u1edbc $A{{l}_{2}}{{O}_{3}}$ \u0111\u01b0\u1ee3c<br\/>$Al+{{O}_{2}}--->2A{{l}_{2}}{{O}_{3}}$<br\/>B\u00ean tr\u00e1i c\u1ea7n 4$Al$ v\u00e0 6$O$ t\u1ee9c 3${{O}_{2}}$, c\u00e1c h\u1ec7 s\u1ed1 4 v\u00e0 3 l\u00e0 th\u00edch h\u1ee3p<br\/>v\u1eady PTHH l\u00e0 $4Al+3{{O}_{2}}\\to 2A{{l}_{2}}{{O}_{3}}$<br\/><br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2111},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"\u0110\u1ec3 \u0111\u1ed1t ch\u00e1y ho\u00e0n to\u00e0n $a$ (g) $Al$ c\u1ea7n d\u00f9ng h\u1ebft $28,8$ g oxi. Ph\u1ea3n \u1ee9ng thu \u0111\u01b0\u1ee3c $m$ (g) $A{{l}_{2}}{{O}_{3}}$. Gi\u00e1 tr\u1ecb $a$ v\u00e0 $m$ l\u00e0 ","select":["A. $a=32,4$ g v\u00e0 $m=61,2$ g ","B. $a=24,3$ g v\u00e0 $m=91,8$ g ","C. $a=16,2$ g v\u00e0 $m=61,2$ g ","D. $a=48,6$ g v\u00e0 $m=183,6$ g "],"hint":"","explain":"<span class='basic_left'>PTHH c\u1ee7a ph\u1ea3n \u1ee9ng l\u00e0 $4Al+3{{O}_{2}}\\to 2A{{l}_{2}}{{O}_{3}}$<br\/>S\u1ed1 mol c\u1ee7a ${{O}_{2}}$ l\u00e0<br\/>$n=\\dfrac{m}{M}=\\dfrac{28,8}{32}=0,9(mol)$<br\/>Theo ph\u01b0\u01a1ng tr\u00ecnh h\u00f3a h\u1ecdc ta c\u00f3:<br\/>C\u1ee9 4 mol $Al$ t\u00e1c d\u1ee5ng v\u1edbi 3 mol ${{O}_{2}}$<br\/>V\u1eady x (mol) $Al$ t\u00e1c d\u1ee5ng v\u1edbi 0,9 (mol) ${{O}_{2}}$<br\/>$\\Rightarrow x=\\dfrac{0,9.4}{3}=1,2(mol)$<br\/>$\\Rightarrow a=1,2.27=32,4$ g<br\/>T\u01b0\u01a1ng t\u1ef1 v\u1edbi $A{{l}_{2}}{{O}_{3}}$ ta c\u00f3 : ${{n}_{A{{l}_{2}}{{O}_{3}}}}=\\dfrac{0,9.2}{3}=0,6(mol)$<br\/>$\\Rightarrow m=0,6.102=61,2$ g<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2112},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"Trong ph\u00f2ng th\u00ed nghi\u1ec7m ng\u01b0\u1eddi ta c\u00f3 th\u1ec3 \u0111i\u1ec1u ch\u1ebf oxi b\u1eb1ng c\u00e1ch nhi\u1ec7t ph\u00e2n kali clorat theo ph\u1ea3n \u1ee9ng : $2KCl{{O}_{3}}\\to 2KCl+3{{O}_{2}}$. Kh\u1ed1i l\u01b0\u1ee3ng $KCl{{O}_{3}}$ c\u1ea7n thi\u1ebft \u0111\u1ec3 \u0111i\u1ec1u ch\u1ebf \u0111\u01b0\u1ee3c $2,4$ g ${{O}_{2}}$ ","select":["A. $7,125$ g ","B. $27,5625$ g ","C. $17,85$ g ","D. $6,125$ g "],"hint":"","explain":"<span class='basic_left'>S\u1ed1 mol c\u1ee7a ${{O}_{2}}$ l\u00e0 $n=\\dfrac{m}{M}=\\dfrac{2.4}{32}=0,075(mol)$<br\/>Theo ph\u01b0\u01a1ng tr\u00ecnh h\u00f3a h\u1ecdc ta c\u00f3:<br\/>C\u1ee9 2 mol $KCl{{O}_{3}}$ tham gia ph\u1ea3n \u1ee9ng s\u1ebd t\u1ea1o ra 3 mol ${{O}_{2}}$<br\/>V\u1eady x (mol) $KCl{{O}_{3}}$ t\u00e1c d\u1ee5ng v\u1edbi 0,075 (mol) ${{O}_{2}}$<br\/>$\\Rightarrow x=\\dfrac{0,075.2}{3}=0,05(mol)$<br\/>$\\Rightarrow$ Kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a $KCl{{O}_{3}}$ l\u00e0 $m=0,05.(39+35,5+16.3)=6,125$ g<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2113},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"Th\u1ec3 t\u00edch c\u1ee7a oxi (\u0111ktc) c\u1ea7n d\u00f9ng \u0111\u1ec3 \u0111\u1ed1t ch\u00e1y h\u1ebft $3,1$ g $P$ thu \u0111\u01b0\u1ee3c $m$ g ${{P}_{2}}{{O}_{5}}$ ","select":["A. $2,8$ l\u00edt ","B. $8,96$ l\u00edt ","C. $2,24$ l\u00edt ","D. $3,36$ l\u00edt "],"hint":"","explain":"<span class='basic_left'>Ph\u01b0\u01a1ng tr\u00ecnh h\u00f3a h\u1ecdc l\u00e0 $4P+5{{O}_{2}}\\to 2{{P}_{2}}{{O}_{5}}$<br\/>S\u1ed1 mol c\u1ee7a $P$ l\u00e0 $n=\\dfrac{m}{M}=\\dfrac{3,1}{31}=0,1(mol)$<br\/>Theo ph\u01b0\u01a1ng tr\u00ecnh h\u00f3a h\u1ecdc ta c\u00f3:<br\/>C\u1ee9 4 mol $P$ t\u00e1c d\u1ee5ng v\u1edbi 5 mol ${{O}_{2}}$<br\/>V\u1eady 0,1 (mol) $P$ t\u00e1c d\u1ee5ng v\u1edbi x (mol) ${{O}_{2}}$<br\/>$\\Rightarrow x=\\dfrac{0,1.5}{4}=0,125(mol)$<br\/>$\\Rightarrow V=0,125.22,4=2,8$ l\u00edt<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2114},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"Bi\u1ebft r\u1eb1ng $2,3$ g m\u1ed9t kim lo\u1ea1i $R$ ( c\u00f3 h\u00f3a tr\u1ecb $I$ ) t\u00e1c d\u1ee5ng v\u1eeba \u0111\u1ee7 v\u1edbi 1,12 l\u00edt kh\u00ed clo \u1edf \u0111ktc theo ph\u01b0\u01a1ng tr\u00ecnh ph\u1ea3n \u1ee9ng sau: $2R+C{{l}_{2}}\\to 2RCl$. Kim lo\u1ea1i $R$ l\u00e0 ","select":["A. $Mg$ ","B. $Cs$ ","C. $K$ ","D. $Na$ "],"hint":"","explain":"<span class='basic_left'>Ta c\u00f3 : ${{n}_{C{{l}_{2}}}}=\\dfrac{1,12}{22,4}=0,05(mol)$<br\/>Theo ph\u01b0\u01a1ng tr\u00ecnh h\u00f3a h\u1ecdc, s\u1ed1 mol $R$ tham gia ph\u1ea3n \u1ee9ng l\u00e0 <br\/>${{n}_{R}}=0,05.2=0,1(mol)$<br\/>$\\Rightarrow$ kh\u1ed1i l\u01b0\u1ee3ng mol c\u1ee7a $R$ l\u00e0 ${{M}_{R}}=\\dfrac{2,3}{0,1}=23$<br\/>V\u1eady $R$ l\u00e0 kim lo\u1ea1i natri $(Na)$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2115},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"Cho kh\u00ed ${{H}_{2}}$ d\u01b0 \u0111i qua \u0111\u1ed3ng $(II)$ oxit n\u00f3ng m\u00e0u \u0111en, ng\u01b0\u1eddi ta thu \u0111\u01b0\u1ee3c $0,32$ g kim lo\u1ea1i $Cu$ m\u00e0u \u0111\u1ecf v\u00e0 h\u01a1i n\u01b0\u1edbc ng\u01b0ng t\u1ee5. S\u1ed1 mol c\u1ee7a $CuO$ tham gia ph\u1ea3n \u1ee9ng l\u00e0 ","select":["A. $0,05$ mol ","B. $0,01$ mol ","C. $0,005$ mol ","D. $0,1$ mol "],"hint":"","explain":"<span class='basic_left'>Ph\u01b0\u01a1ng tr\u00ecnh h\u00f3a h\u1ecdc l\u00e0 ${{H}_{2}}+CuO\\to Cu+{{H}_{2}}O$<br\/>Ta c\u00f3 : ${{n}_{Cu}}=\\dfrac{0,32}{64}=0,005$ mol<br\/>Theo ph\u01b0\u01a1ng tr\u00ecnh ph\u1ea3n \u1ee9ng, s\u1ed1 mol $CuO$ tham gia ph\u1ea3n \u1ee9ng l\u00e0 ${{n}_{CuO}}=0,005$ mol<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2116},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"\u0110\u1ed1t n\u00f3ng $1,35$ g b\u1ed9t $Al$ trong kh\u00ed clo, ng\u01b0\u1eddi ta thu \u0111\u01b0\u1ee3c $6,675$ g nh\u00f4m clorua. C\u00f4ng th\u1ee9c h\u00f3a h\u1ecdc \u0111\u01a1n gi\u1ea3n c\u1ee7a nh\u00f4m clorua l\u00e0 ","select":["A. $A{{l}_{3}}Cl$ ","B. $Al{{Cl}_{3}}$ ","C. $AlCl$ ","D. $A{{l}_{2}}{{Cl}_{3}}$ "],"hint":"","explain":"<span class='basic_left'>S\u1ed1 mol c\u1ee7a $Al$ tham gia ph\u1ea3n \u1ee9ng l\u00e0 $n=\\dfrac{1,35}{27}=0,05(mol)$<br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh lu\u1eadt b\u1ea3o to\u00e0n kh\u1ed1i l\u01b0\u1ee3ng<br\/>$\\Rightarrow {{m}_{Clo}}=6,675-1,35=5,325(g)$<br\/>S\u1ed1 mol c\u1ee7a clo tham gia k\u1ebft h\u1ee3p v\u1edbi $Al$ l\u00e0 $n=\\dfrac{5,325}{35,5}=0,15(mol)$<br\/>G\u1ecdi c\u00f4ng th\u1ee9c c\u1ee7a nh\u00f4m clorua d\u1ea1ng $A{{l}_{x}}{{Cl}_{y}}$<br\/>Trong 1 mol h\u1ee3p ch\u1ea5t t\u1ef7 l\u1ec7 s\u1ed1 nguy\u00ean t\u1eed c\u0169ng l\u00e0 t\u1ef7 l\u1ec7 v\u1ec1 s\u1ed1 mol nguy\u00ean t\u1eed<br\/>$\\Rightarrow \\dfrac{x}{y}=\\dfrac{{{n}_{Al}}}{{{n}_{Cl}}}=\\dfrac{0,05}{0,15}=\\dfrac{1}{3}$<br\/>V\u1eady CTHH l\u00e0 $Al{{Cl}_{3}}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2117},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"Cho $32$ g $CuO$ t\u00e1c d\u1ee5ng v\u1eeba \u0111\u1ee7 v\u1edbi ${{H}_{2}}S{{O}_{4}}$ thu \u0111\u01b0\u1ee3c mu\u1ed1i $CuS{{O}_{4}}$ v\u00e0 ${{H}_{2}}O$. S\u1ed1 ph\u00e2n t\u1eed n\u01b0\u1edbc t\u1ea1o th\u00e0nh l\u00e0 ","select":["A. ${{2,4.10}^{23}}$ ","B. ${{2,6.10}^{23}}$ ","C. ${{3,4.10}^{23}}$ ","D. ${{2,4.10}^{22}}$ "],"hint":"","explain":"<span class='basic_left'>Ph\u01b0\u01a1ng tr\u00ecnh h\u00f3a h\u1ecdc l\u00e0 $CuO+{{H}_{2}}S{{O}_{4}}\\to CuS{{O}_{4}}+{{H}_{2}}O$<br\/>S\u1ed1 mol c\u1ee7a $CuO$ l\u00e0 ${{n}_{CuO}}=\\dfrac{32}{80}=0,4(mol)$<br\/>Theo ph\u01b0\u01a1ng tr\u00ecnh ph\u1ea3n \u1ee9ng, s\u1ed1 mol ${{H}_{2}}O$ l\u00e0 0,4 mol<br\/>$\\Rightarrow$ S\u1ed1 ph\u00e2n t\u1eed n\u01b0\u1edbc l\u00e0 ${{0,4.6.10}^{23}}$ = ${{2,4.10}^{23}}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2118},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"C\u00e2y xanh quang h\u1ee3p theo ph\u01b0\u01a1ng tr\u00ecnh : $nC{{O}_{2}}+5n{{H}_{2}}O\\to {{({{C}_{6}}{{H}_{10}}{{O}_{5}})}_{n}}+6n{{O}_{2}}$. Kh\u1ed1i l\u01b0\u1ee3ng tinh b\u1ed9t ${{({{C}_{6}}{{H}_{10}}{{O}_{5}})}_{n}}$ thu \u0111\u01b0\u1ee3c n\u1ebfu kh\u1ed1i l\u01b0\u1ee3ng ${{H}_{2}}O$ ti\u00eau th\u1ee5 l\u00e0 $9$ t\u1ea5n ","select":["A. $19,9$ t\u1ea5n ","B. $17,5$ t\u1ea5n ","C. $18,9$ t\u1ea5n ","D. $16,2$ t\u1ea5n "],"hint":"","explain":"<span class='basic_left'>Ta c\u00f3 : $n=\\dfrac{{{9.10}^{6}}}{18}={{0,5.10}^{6}}(mol)$<br\/>Theo ph\u01b0\u01a1ng tr\u00ecnh h\u00f3a h\u1ecdc, s\u1ed1 mol ${{({{C}_{6}}{{H}_{10}}{{O}_{5}})}_{n}}$ l\u00e0 <br\/>${{n}_{{{({{C}_{6}}{{H}_{10}}{{O}_{5}})}_{n}}}}=\\dfrac{{{0,5.10}^{6}}}{5n}=\\dfrac{{{0,1.10}^{6}}}{n}(mol)$<br\/>$\\Rightarrow {{m}_{{{({{C}_{6}}{{H}_{10}}{{O}_{5}})}_{n}}}}=\\dfrac{{{0,1.10}^{6}}}{n}.162.n={{16,2.10}^{6}}(g)=16,2$ t\u1ea5n<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2119}],"lesson":{"save":0,"level":2}}