醋酸和氢氧化钡反应化学方程式 氢氧化钡和醋酸怎么反应，方程式是怎么样的
\u6c22\u6c27\u5316\u94a1\u548c\u918b\u9178\u53cd\u5e94\u7684\u79bb\u5b50\u65b9\u7a0b\u5f0f\u5230\u5e95\u662f\u54ea\u4e00\u79cd???\u4e2d\u548c\u53cd\u5e94 Ba(OH)2+2CH3COOH=====Ba(CH3COO)2+2H2O
\u6c22\u6c27\u5316\u94a1\u548c\u918b\u9178\u53cd\u5e94\u4f1a\u751f\u6210\u918b\u9178\u94a1\u548c\u6c34
\u65b9\u7a0b\u5f0f\uff1a2CH3COOH Ba\uff08OH\uff092=\uff08CH3COOH\uff092Ba 2H2O
=Ba(CH3COO)2↓+2H2O
绛旓細(25)纭鍜屾阿姘у寲閽婧舵恫鍙嶅簲:Si+2OH-+H2O===SiO3-+2H2鈫戜竷銆侀噾灞炰笌鐩愬弽搴(26)閾佸拰纭吀閾滄憾娑插弽搴:Fe+Cu2+===Fe2++Cu(27)閿屼笌纭吀閾滄憾娑插弽搴::Zn+Cu2+===Zn2++Cu(28)閾滀笌纭濋吀閾舵憾娑插弽搴:Cu+2Ag+===Cu2++2Ag(29)閾濆拰纭吀閾滄憾娑插弽搴:2Al+3Cu2+===2Al3++3Cu(30)閾佸拰姘寲閾佹憾娑插弽搴:...
绛旓細鐩愰吀鍜岀⒊閰搁挋 CaCO3+2H+===H2O+CO2+Ca2+ 姘㈡哀鍖栭挕鍜岀█纭吀 Ba2+ +2OH-+2H= +SO4 2-==BaSO4!+2H2O 姘ㄦ按鍜閱嬮吀 CH3COOH+NH3*H2O==CH3COO- +NH4+ +|H2O 姘寲浜氶搧鍜屾隘姘旂殑绂诲瓙鏂圭▼寮 2Fe2+ +Cl2===2Fe3+ +2Cl-
绛旓細姹姘㈡哀鍖栭挕鍜岀～閰,姘㈡哀鍖栭捑鍜閱嬮吀,纰抽吀閽犲拰纭濋吀,纰抽吀閽欏拰绋鐩愰吀鐨勭瀛鏂圭▼寮,鏈濂芥墜鍐,璋簡 鎴戞潵绛 2涓洖绛 #鐑# 浣犲彂鏈嬪弸鍦堜細浣跨敤閮ㄥ垎浜哄彲瑙佸姛鑳藉悧?CC鐨灏忓垵楂樿祫鏂欓叿 2015-11-02 路 鐭ラ亾鍚堜紮浜烘暀鑲茶瀹 CC鐨勫皬鍒濋珮璧勬枡閰 鐭ラ亾鍚堜紮浜烘暀鑲茶瀹 閲囩撼鏁:1564 鑾疯禐鏁:17629 楠ㄥ共鍖栧鏁欏笀,瀛︾敓...
绛旓細濡:閱嬮吀涓庢阿姘у寲閽鍙嶅簲 鍖栧鏂圭▼寮:CH3COOH + NaOH = CH3COONa + H2O 绂诲瓙鏂圭▼寮: CH3COOH + OH- = CH3COO- + H2O 杩欎釜鍙嶅簲鏄急閰镐笌寮虹⒈鐨勫弽搴.閱嬮吀涓嶈兘鎷.鍙︿竴鐢熸垚娌夋穩鐨,濡姘㈡哀鍖栭挕涓庣～閰哥殑鍙嶅簲:鍖栧鏂圭▼寮:Ba(OH)2 + H2SO4 = 2H2O + BaSO4鈫 绂诲瓙鏂圭▼寮:Ba2+ + 2OH- + 2H+ + ...
绛旓細姘㈡哀鍖栭挕涓轰簩鍏冨己纰憋紝浜岃呭弽搴旂敓鎴愮～閰搁挕娌夋穩鍜屾按锛鍙嶅簲鐨绂诲瓙鏂圭▼寮涓猴細2H + +2OH - +SO 4 2- +Ba 2+ =BaSO 4 鈫+2H 2 O锛屾晠绛旀涓猴細Ba 2+ +2OH - +2H + +SO 4 2- 鈺怋aSO 4 鈫+2H 2 O锛4锛夌⒊閰告阿閽犱负鍙憾鎬х洂锛屾阿姘у寲閽犱负涓鍏冨己纰憋紝浜岃呭弽搴旂敓鎴愮⒊閰搁挔鍜屾按锛屽弽搴旂殑绂诲瓙...
绛旓細鍖栧鏂圭▼寮鍙嶆槧鐨勬槸瀹㈣浜嬪疄銆傚洜姝や功鍐欏寲瀛︽柟绋嬪紡瑕侀伒瀹堜袱涓師鍒:涓鏄繀椤讳互瀹㈣浜嬪疄涓哄熀纭;浜屾槸瑕侀伒瀹堣川閲忓畧鎭掑畾寰嬨 1銆侀挔涓庢按鍙嶅簲:2Na+2H2O=2Na++2OH-+H2鈫 2銆佺閰搁摱涓庣洂閰稿強鍙憾鎬х洂閰哥洂婧舵恫:Ag++Cl-=AgCl鈫 3銆佽繃姘у寲閽犱笌姘村弽搴:2Na2O+2H2O=4Na++4OH-+O2鈫 4銆佺⒊閰姘鐩婧舵恫涓寮洪吀婧舵恫娣峰悎:HCO3...
绛旓細(1)姘㈡哀鍖栭挕婧舵恫闃寸瀛愮鍙 OH^- (2)鍖栧寮CH3COOH (3)+6浠
绛旓細27銆佺～閰告憾娑璺熸阿姘у寲閽℃憾娑鍙嶅簲:Ba2++2OH鈥+2H++SO42鈥=BaSO4鈫+2H2O28銆佺～閰告阿閽婧舵恫涓庢阿姘у寲閽″弽搴鑷充腑鎬:2H++SO42鈥+2OH鈥+Ba2+=2H2O+BaSO4鈫29銆佺～閰告阿閽犳憾娑蹭笌姘㈡哀鍖栭挕鍙嶅簲鑷崇～閰告牴瀹屽叏娌夋穩:H++SO42鈥+OH鈥+Ba2+=BaSO4鈫+H2O30銆佺～閰搁摑婧舵恫涓姞鍏ヨ繃閲忔阿姘у寲閽℃憾娑:2Al3++3SO42鈥+8OH...
绛旓細纭吀闀佹憾娑璺熸阿姘у寲閽℃憾娑鍙嶅簲: Ba2++2OH鈥+Mg2++SO42鈥=BaSO4鈫+Mg(OH)2鈫 27銆佺～閰告憾娑茶窡姘㈡哀鍖栭挕婧舵恫鍙嶅簲:Ba2++2OH鈥+2H++SO42鈥=BaSO4鈫+2H2O 28銆佺～閰告阿閽婧舵恫涓庢阿姘у寲閽″弽搴鑷充腑鎬:2H++SO42鈥+2OH鈥+Ba2+=2H2O+BaSO4鈫 29銆佺～閰告阿閽犳憾娑蹭笌姘㈡哀鍖栭挕鍙嶅簲鑷崇～閰告牴瀹屽叏娌夋穩: H++SO42鈥...
绛旓細锛1锛夆憼宸茬煡绋寮洪吀銆佺█寮虹⒈鍙嶅簲鐢熸垚1mol娑叉佹按鏃舵斁鍑57.3kJ鐨勭儹閲忥紝绋鐩愰吀鍜屾阿姘у寲閽绋婧舵恫閮芥槸寮洪吀鍜屽己纰辩殑绋婧舵恫锛屽垯鍙嶅簲鐨勭儹鍖栧鏂圭▼寮涓12Ba锛圤H锛2锛坅q锛+HCl 锛坅q锛夆晲12BaCl2锛坅q锛+H2O鈻矵=-57.3 kJ/mol锛屾晠绛旀涓猴細12Ba锛圤H锛2锛坅q锛+HCl 锛坅q锛夆晲12BaCl2锛坅q锛+H2O鈻矵=-57.3...
