{"common":{"save":0,"post_id":"8541","level":2,"total":10,"point":10,"point_extra":0},"segment":[{"id":"7649","post_id":"8541","mon_id":"1159539","chapter_id":"1159541","question":"<p>Cho PTHH: Ba + 2HCl → BaCl<span class=\"math-tex\">$_{2}$<\/span> + H<span class=\"math-tex\">$_{2}$<\/span>. \u0110\u1ec3 thu d\u01b0\u1ee3c 4,16 g BaCl<span class=\"math-tex\">$_{2}$<\/span> c\u1ea7n bao nhiêu mol HCl?<\/p>","options":["A. 0,04 mol","B. 0,01 mol","C. 0,02 mol","D. 0,5 mol"],"correct":"1","level":"2","hint":"<p>Hãy tính mol BaCl<span class=\"math-tex\">$_{2}$<\/span> t\u1eeb kh\u1ed1i l\u01b0\u1ee3ng và s\u1eed d\u1ee5ng t\u1ec9 l\u1ec7 ph\u1ea3n \u1ee9ng \u0111\u1ec3 xác \u0111\u1ecbnh s\u1ed1 mol HCl c\u1ea7n thi\u1ebft.<\/p>","answer":"<p>\u0110áp án \u0111úng: <strong>A. 0,04 mol<\/strong><\/p><p>n<span class=\"math-tex\">$_{BaCl_{2}}$<\/span> = 4,16 \/ 208 = 0,02 (mol)<\/p><p>theo PTHH, ta có: n<span class=\"math-tex\">$_{HCl}$<\/span> = 0,04 (mol)<\/p>","type":"choose","extra_type":"classic","time":"0","user_id":"139","test":"0","date":"2025-02-26 10:32:34","option_type":"txt","len":1},{"id":"7650","post_id":"8541","mon_id":"1159539","chapter_id":"1159541","question":"<p>Nung 6,72 g Fe trong không khí thu \u0111\u01b0\u1ee3c s\u1eaft (II) oxit. Tính mFeO và VO2?<\/p>","options":["A. 1,344g và 0,684 lít","B. 2,688 lít và 0,864g","C. 1,344 lít và 8,64g","D. 8,64g và 2,234 ml"],"correct":"3","level":"2","hint":"<p>Hãy tính s\u1ed1 mol Fe tham gia ph\u1ea3n \u1ee9ng và s\u1eed d\u1ee5ng \u0111\u1ecbnh lý b\u1ea3o toàn kh\u1ed1i l\u01b0\u1ee3ng \u0111\u1ec3 tìm mFeO và th\u1ec3 tích O2.<\/p>","answer":"<p>\u0110áp án \u0111úng: <strong>C. 1,344 lít và 8,64g<\/strong><\/p><p>nFe = 6,72 \/ 56 = 0,12 (mol)<\/p><p>PTHH: 2Fe + O<span class=\"math-tex\">$_{2}$<\/span> -> 2FeO<\/p><p>Theo PTHH: n<span class=\"math-tex\">$_{O_{2}}$<\/span>= 0,06 (mol)<\/p><p>=> V<span class=\"math-tex\">$_{O_{2}}$<\/span> = 0,06.22,4 = 1,344 (lít)<\/p><p>Theo PTHH: n<span class=\"math-tex\">$_{FeO}$<\/span> = 0,12 (mol)<\/p><p>=> m<span class=\"math-tex\">$_{FeO}$<\/span> = 0,12.(56+16) = 8,64 (gam)<\/p>","type":"choose","extra_type":"classic","time":"0","user_id":"139","test":"0","date":"2025-02-26 10:37:01","option_type":"txt","len":2},{"id":"7651","post_id":"8541","mon_id":"1159539","chapter_id":"1159541","question":"<p>Cho 2,7 g nhôm tác d\u1ee5ng v\u1edbi oxi, sau ph\u1ea3n \u1ee9ng thu \u0111\u01b0\u1ee3c bao nhiêu gam nhôm oxit?<\/p>","options":["A. 1,02 gam","B. 20,4 gam","C. 10,2 gam","D. 5,1 gam"],"correct":"4","level":"2","hint":"<p>Hãy nh\u1edb r\u1eb1ng ph\u1ea3n \u1ee9ng gi\u1eefa nhôm và oxi t\u1ea1o ra nhôm oxit (Al\u2082O\u2083). Dùng \u0111\u1ecbnh lý b\u1ea3o toàn kh\u1ed1i l\u01b0\u1ee3ng và tính toán s\u1ed1 mol c\u1ee7a nhôm tham gia ph\u1ea3n \u1ee9ng.<\/p>","answer":"<p>\u0110áp án \u0111úng: <strong>D. 5,1 gam<\/strong><\/p><p>n<span class=\"math-tex\">$_{Al}$<\/span> = 2,7 \/ 27 = 0,1 (mol)<\/p><p>PTHH: 4Al +3O<span class=\"math-tex\">$_{2}$<\/span> -> 2Al<span class=\"math-tex\">$_{2}$<\/span>O<span class=\"math-tex\">$_{3}$<\/span><\/p><p>Theo PTHH: n<span class=\"math-tex\">$_{Al_{2}O_{3}}$<\/span> = 0,05 (mol)<\/p><p>=> m<span class=\"math-tex\">$_{Al_{2}O_{3}}$<\/span> = 0,05.102 = 5,1 (gam)<\/p>","type":"choose","extra_type":"classic","time":"0","user_id":"139","test":"0","date":"2025-02-26 10:40:45","option_type":"txt","len":1}]}
