一道化学题
\u4e00\u9053\u5316\u5b66\u9898\uff0c\u6025\uff0c\u5728\u7ebf\u7b49\u8be5\u70c3\u7684\u6469\u5c14\u8d28\u91cf\u4e3a\uff1a3.75g/L*22.4L/mol=84g/mol
n(CO2)=5.8/44=0.13182mol
n(H2O)=2.16/18=0.12mol
n(C):C(H)=1:2 \uff08\u53d6\u6574\u7ea6\u7b49\u4e8e1\uff1a2\uff0c\u78b3\u6c22\u5747\u6765\u81ea\u4e8e\u70c3\uff09\u5206\u5b50\u5f0f\u4e3a\uff1a(CH2)n 14n=84 \u5f97n=6
\u6545\u5206\u5b50\u5f0f\u4e3aC6H12\uff0c\u53ef\u80fd\u7684\u7ed3\u6784\u7b80\u5f0f\u8f83\u591a\uff0c\u73af\u72b6\u4e3a\uff1a\u73af\u5df1\u70f7\uff1b\u76f4\u94fe\u578b\u4e3a\uff1a
CH2\uff1dCHCH2CH2CH2CH3\u3001CH3CH\uff1dCHCH2CH2CH3\u3001CH3CH2CH\uff1dCHCH2CH3\uff1b\u652f\u94fe\u578b\u8f83\u591a\uff0c\u4f60\u53ef\u4ee5\u81ea\u5df1\u6392\u4e00\u4e0b\u3002
\u6b64\u5916\uff0c\u6b64\u9898\u4e2d\u524d\u9762\u7b2c\u4e00\u6b65\u6c42\u51fa\u70c3\u7684\u6469\u5c14\u8d28\u91cf\u4e3a84\uff0c\u518d\u6839\u636e\u8d28\u91cf\u53ef\u4ee5\u6c42\u51fa\u70c3\u7684\u7269\u8d28\u7684\u91cf\u4e3a1.68/84=0.02mol.\u518d\u6839\u636e\u751f\u6210CO2\u7684\u7269\u8d28\u7684\u91cf0.132182mol\u548c\u751f\u6210H2O\u7269\u8d28\u7684\u91cf0.12mol.\u53ef\u6c42\u51faC\u548c\u6c22\u7684\u539f\u5b50\u6570\u5206\u522b\u4e3a6\u548c12\uff08\u53d6\u6574\u6570\uff09
1\u3001\u751f\u6210CO2\u4f7f\u6db2\u6001\u8d28\u91cf\u51cf\u5c11121-50-11-64.4\uff1d4.4g\uff0c\u78b3\u9178\u94a0\u7684\u8d28\u91cf\u4e3a\uff1a4.4*106/44\uff1d10.6g \u7eaf\u5ea6\u4e3a10.6*100%/11=96.36%
2.Na2SO4%=(4.4*142/44)*100%/(121-(11-10.6))=11.77%\uff1b NaCl%=\uff0811-10.6\uff09*100%/(121-14.2)=0.37%
从题意可以知道A,B,C分别为戊酸，戊醇，戊酸戊酯
戊酸有四种异构体
CH3CH2CH2CH2COOH,
CH3CH2CH(CH3)COOH,
(CH3)3CCOOH,
(CH3)2CHCH2COOH
对应的有四种醇，CH3CH2CH2CH2CH2OH;CH3CH2CH(CH3)CH2OH;(CH3)3CCH2OH;(CH3)2CHCH2CH2OH因此能生成四种酯的异构体
A、B可写成：
A：C4H9COOH、B：C4H9CH2OH
形成的酯可写成C4H9COOCH2C4H9
-C4H9是丁基，有四种结构
①-CH2CH2CH2CH3
②-CHCH2CH3
|
CH3
③-CH2CHCH3
|
CH3
④ CH3
|
-CCH3
|
CH3
对应的分子式为C10H20O2的酯为4种，分别为
①CH3CH2CH2CH2-COO-CH2CH2CH2CH2CH3
②(CH3)2CHCH2-COO-CH2CH2CH (CH3)2
③CH3CH2CH(CH3)-COO-CH2CH(CH3)CH2CH3
④(CH3)3C-COO-CH2C(CH3)3
不对的就删了
D.4种 HCOOCH2CH2CH2CH3 CH3COOCH2CH2CH3 CH3CH2COOCH2CH3 CH3CH2CH2COOCH3
绛旓細鎬ワ紝涓閬撳寲瀛﹂锛屽寲瀛﹂珮鎵嬫潵鍟婏紒 锛1锛塁u+2H2SO4(娴)==(鍔犵儹)CuSO4+2H2O+SO2(姘斾綋) Fe+CuSO4==FeSO4+Cu Fe+H2SO4==FeSO4+H2(姘斾綋) 锛2锛夌敱棰橈細 n(SO2)=13.44L/22.4(L/mol)=0.4mol n(H2)=0.4mol 鍙堬細 SO2鈥2H2SO4 0.4mol鈥0.8mol H2鈥擧2SO4 0.4m...
绛旓細1銆佹皵浣揂鏄 H2鍜孋O鐨勬贩鍚堟皵浣撱傞粦鑹茬矇鏈槸CuO銆傘愯В鏋愩戠敳璇曠鐜拌薄鏄孩鑹插浐浣擄紝鍏堕粦鑹茬矇鏈槸绾噣鐗╋紝鎵浠ョ孩鑹插浐浣撴槸Cu锛岄粦鑹茬矇鏈嚜鐒跺氨鏄疌uO銆備箼瑁呯疆鍙樹负钃濊壊锛屾槸鏃犳按CuSO4鍜孒2O鍙嶅簲鐢熸垚CuSO4路5H2O锛屾墍浠ユ皵浣揂鍚湁H2 鑰屼笝瑁呯疆娴戞祳锛屽垯鏄敓鎴愪簡CO2姘斾綋鍜孋a(OH)2鍙嶅簲鐢熸垚浜嗘矇娣锛屾墍浠ユ皵浣揂鍚湁CO 2銆...
绛旓細锛1锛夊洜涓虹⒊閰搁挋鍜岀洂閰稿弽搴斿悗浼氫骇鐢熶簩姘у寲纰虫皵浣擄紝鎵浠ョ儳鏉腑鐗╄川鎬昏川閲忎細鍑忓皯锛屾牴鎹川閲忓畧鎭掑畾寰嬪彲鐭ワ紝鐑ф澂涓墿璐ㄦ昏川閲忕殑鍑忓皯閲忓氨鏄敓鎴愮殑浜屾哀鍖栫⒊鐨勮川閲忥紝鏍规嵁鍖栧鏂圭▼寮忓埄鐢ㄤ簩姘у寲纰崇殑璐ㄩ噺鍙緱鍒扮⒊閰搁挋鐨勮川閲忥紱锛2锛夋牴鎹⒊閰搁挋鍜岀洂閰稿弽搴旂殑鍖栧鏂圭▼寮忥紝鍒╃敤浜屾哀鍖栫⒊鐨勮川閲忓彲姹傚嚭鐢熸垚鐨勬隘鍖栭挋鐨勮川閲忥紝杩涜屾眰鍑...
绛旓細姹涓閬撳寲瀛﹂鐨勮В绛旇繃绋 n(AgNO3)=0.2*5=1 m mol n(AgCl)=143.5/143.5=1 m mol n(Cl)=1*500/10=50 m mol=0.05 mol n(P)=0.31/31=0.01 mol x=n(Cl)/n(P)=5.涓閬撳寲瀛﹂鐨勮В绛 鈫撯啈 鈫 鈫撯啈 鈫 鈫 娓哥爜鍦ㄦ爣灏虹殑浣嶇疆=0.8g 璁颁綇鐮濈爜鐨勫彇鐢ㄥ師鍒欏氨鍙互浜 ...
绛旓細鍒嗘瀽 锛1锛夋牴鎹浘鍍忓彲鐭ワ細鏇茬嚎鐨勬姌鐐规墍瀵瑰簲鐨勭⒊閰搁挔婧舵恫鐨勮川閲忓嵆鎭板ソ瀹屽叏鍙嶅簲鏃舵墍娑堣楃殑璐ㄩ噺锛涳紙2锛夋牴鎹弽搴旂殑鍖栧鏂圭▼寮忥紝鍒╃敤鎭板ソ瀹屽叏鍙嶅簲鏃舵秷鑰楃⒊閰搁挔鐨勮川閲忔垨鐢熸垚娌夋穩鐨勮川閲忥紝璁＄畻涓庣⒊閰搁挔鍙戠敓鍙嶅簲鐨勬隘鍖栭挕鐨勮川閲忥紝鐢辨隘鍖栭挕璐ㄩ噺涓庝骇鍝佽川閲忓埄鐢ㄨ川閲忓垎鏁拌绠楀叕寮忚绠楀嚭浜у搧涓隘鍖栭挕鐨勮川閲忓垎鏁板嵆鍙紱锛3锛...
绛旓細1.Zn+CuO=Cu+ZnO;2.Zn+CO2涓嶅彂鐢鍖栧鍙嶅簲锛3.Zn+H2SO4=ZnSO4+H2鈫;4.Zn+2HCl=ZnCl2+H2鈫;5.Zn+CuSO4=ZnSO4+Cu;6.Zn+2AgNO3=Zn锛圢O3)2+Ag;7.Zn+BaCl2涓嶅彂鐢熷寲瀛﹀弽搴旓紝閿屽厓绱犳病鏈夐挕鍏冪礌娲绘臣锛8.Zn+CuCO3=ZnCO3+Cu锛9.Zn+NaOH涓嶅彂鐢熷寲瀛﹀弽搴旓紱10.Zn+Al锛圤H锛3涓嶅彂鐢熷寲瀛﹀弽搴斻備簩銆...
绛旓細鐕冪儳閫氬紡锛欳xH(2x+2) + (3x/2+1/2)O2 == xCO2 + (x+1)H2O 0.1molA瀹屽叏鐕冪儳鐨勪骇鐗╅氳繃纰辩煶鐏帮紝纰辩煶鐏板閲20.4g銆傚嵆锛0.1*[44x + 18(x+1)] = 20.4锛岃В寰梮 = 3 鎵浠ョ儍A灞炰簬銆愮兎鐑冦戯紝鍒嗗瓙寮忋怌3H8銆(2)鐕冪儳鍚庯紝鎭㈠鍒板師鎬侊紝鍒欑敓鎴愮殑姘存槸姘旀併傚畬鍏ㄥ弽搴旂敓鎴怌O2鍜孒2O鐨勬柟绋嬪紡...
绛旓細2Na2O2+2CO2===2Na2CO3+O2 4*24*50L*1.5g/L=7200g=7.2kg 2Na2O2+2CO2===2Na2CO3+O2 156 32 x 7.2kg 156/x=32/7.2kg x=35.1kg
绛旓細A銆侀噾灞為捑鐢变簬澶椿娉硷紝鍏堜笌姘村弽搴斾簡锛2K + 2H2O = 2KOH + H2鈫戯紝鐒跺悗锛孼nSO4 + 2KOH = Zn(OH)2鈫 + K2SO4锛屾渶鍚庡緱鍒扮殑鏄痁n(OH)2娌夋穩锛岃屼笉鏄痁n鍗曡川銆侭銆丗e + H2SO4 = FeSO4 + H2鈫 56 2 褰撴憾瑙56鍏嬮搧鍚屾椂鏀惧嚭2鍏嬫阿姘旓紝婧舵恫澧為噸56g - 2g = 54g銆侰銆佸湪绌烘皵涓姞鐑摑锛屽笇...
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