询问一下关于化学最后一道计算题的问题？ 怎么做化学卷子的最后一道计算题?
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1\u3001\u91cd\u5927\u79d1\u6280\u6210\u679c\u7c7b\uff1a
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2\u3001\u91cd\u5927\u5b89\u5168\u4e8b\u6545\u7c7b\uff1a
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3\u3001\u6ce8\u91cd\u4e0e\u6240\u6559\u73ed\u7ea7\u73ed\u4e3b\u4efb\u8001\u5e08\u7684\u534f\u8c03\u6c9f\u901a\u3002\u5e38\u8a00\u9053\uff0c\u201c\u4e00\u82b1\u72ec\u653e\u4e0d\u662f\u6625\u201d\uff0c\u6293\u4f18\u79c0\u7387\u548c\u5e73\u5747\u5206\uff0c\u9700\u8981\u4e0e\u73ed\u4e3b\u4efb\u6559\u5e08\u6b65\u8c03\u4e00\u81f4\uff0c\u76f8\u4e92\u5c0a\u91cd\u548c\u76f8\u4e92\u7406\u89e3\uff0c\u5f62\u6210\u5408\u529b\uff0c\u5728\u6709\u9650\u7684\u65f6\u95f4\u5185\u6293\u51fa\u6210\u7ee9\uff0c\u6293\u51fa\u8d28\u91cf\u3002
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要注意一些条件，不如它说稀盐酸多少克，那么这个是不可以代入计算的。代入计算的必须是纯净的~~比如说：碳酸钙多少克
然后通过比例求解
最后计算质量分数就好了
例题：测定某盐酸质量分数，取10g碳酸钙，滴加足量该盐酸，反应停止时，测得使用盐酸71g
分析：盐酸是混合物，所以不能用来计算。碳酸钙则可
解：设HCl质量为x
CaCO3+2HCl=CaCl2+H2O+CO2↑
100 71
10g x
100/10g=71/x
x=7.1g
HCl%=7.1g/71g*100%=10%
答：………………
给你推荐的复习内容：一是表格题，表格题要看懂反应的量的关系，谁过量，谁少量，谁少量用谁算。二是图像题，口诀：一轴二线三点。要注意起点、拐点中终点
中考化学的最后一题一般都不难，大部分都是用化学方程式解某个量，只要注意方程式的一些必要点如条件不能忘等就可以了
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绛旓細鏈鍚庝竴棰閮戒笉闅撅紝鍙鎵惧埌鍖栧鏂圭▼鍚勯噺鐨勫叧绯诲嵆鍙 瑕佹敞鎰忎竴浜涙潯浠讹紝涓嶅瀹冭绋鐩愰吀澶氬皯鍏嬶紝閭ｄ箞杩欎釜鏄笉鍙互浠ｅ叆璁＄畻鐨勩備唬鍏ヨ绠楃殑蹇呴』鏄函鍑鐨剘~姣斿璇达細纰抽吀閽欏灏戝厠 鐒跺悗閫氳繃姣斾緥姹傝В 鏈鍚庤绠璐ㄩ噺鍒嗘暟灏卞ソ浜 渚嬮锛氭祴瀹氭煇鐩愰吀璐ㄩ噺鍒嗘暟锛屽彇10g纰抽吀閽欙紝婊村姞瓒抽噺璇ョ洂閰革紝鍙嶅簲鍋滄鏃讹紝娴嬪緱浣跨敤鐩愰吀71g ...
绛旓細鍘嬭酱棰 鍦ㄦ垜浠殑涓鍖栧涓紝鏈鍚庝竴閬棰橀兘鏄竴閬撶患鍚璁＄畻棰銆傝繖閬撻鍒嗗间笉浣庯紝棰樼洰鐪嬭捣鏉ユ湁鐐归暱锛岃绠楃殑杩囩▼涔熶笉涓瀹氭槸涓姝ュ埌浣嶏紝灏ゅ叾鍋氬埌鏈鍚庢椂闂翠篃姣旇緝绱ц揩锛屾墍浠ュぇ瀹堕兘浼氳寰楄繖閬撻涓嶇畝鍗曘傚叾瀹炲苟涓嶆槸杩欐牱鐨勶紝鍙嶈鍚勫湴銆佸巻骞寸殑涓冮涓紝杩欓亾棰樿冪殑閮戒笉闅撅紝鍙浣犳妸鏁欐潗涓渶鍩烘湰鐨勮绠楀叕寮忔帉鎻★紝...
绛旓細H:O=8锛16*4=1锛8 10*85%*23/12*5+8+14+16*4+23= 26.2H2O=2H2+O2 36 4 x 3 36/4=x/3 x=27 鑳界湅鏄庣櫧鍚 璇烽噰绾
绛旓細鐢辩涓闂鐭,n(NaOH)=0.15mol 鍦ㄧ浜岄棶杩涜鍒嗙被璁ㄨ 1 鑻ョ敓鎴愮殑鐧借壊鍥轰綋涓篘a2CO3鍜孨aHCO3涓殑涓绉嶆垨涓ょ,璁緉(Na2CO3)=x,n(NaHCO3)=a,n(CO2)=n,鏈22.4n=V,鍙圴/N=0.1,鏁匩=224n 鎵浠+a=n,2x+a=0.15,106x+84a=N=224n 瑙ｅ緱a涓鸿礋鏁,鑸 2 鑻ョ敓鎴愮殑鐧借壊鍥轰綋涓篘a2CO3锛堝繀鏈夛級...
绛旓細杩欐槸姣斾緥鐨璁＄畻锛屾槸鍙互杩欐牱璁＄畻鐨勩備綘涔熷彲浠ユ寜鐓у父瑙勭殑璁＄畻鏂规硶锛80脳1.1 = 44x 88 = 44x 鎵浠ワ紝x = 2 鍏跺疄锛屽彧瑕鍖栧鏂圭▼寮忓垪瀵逛簡锛岃绠楄繃绋嬫棤闈炲氨鏄洓鍒欒繍绠楋紙鍔犮佸噺銆佷箻銆侀櫎锛夈傚彧瑕佷笉杩濊儗杩愮畻瑙勫緥锛堝厛涔橀櫎鍚庡姞鍑忋佷箻娉曞垎閰嶅緥銆佷氦鎹㈠緥绛夌瓑锛夛紝璁＄畻缁撴灉搴旇鏄竴鏍风殑銆
绛旓細鏈鍚庝竴閬撹绠楅锛屽鏄绠楀弽搴斿悗婧惰川鐨勮川閲忓垎鏁颁互鍙婂姞鍏ユ憾娑蹭腑婧惰川鐨勮川閲忓垎鏁般傛憾璐ㄧ殑璐ㄩ噺鍒嗘暟锛屼竴鑸槸鍏堟眰鍑烘憾璐ㄧ殑璐ㄩ噺鍜屾憾娑茬殑璐ㄩ噺锛屽啀鐢ㄦ憾璐ㄨ川閲忔瘮涓婃憾娑茶川閲徝100锛呫傚叾涓紝姹傛憾娑茬殑璐ㄩ噺鏄湁涓瀹氭妧宸х殑銆備竴鑸负涓ょ锛堟垨澶氱锛夋憾娑插弽搴旓紝鍙渶灏嗕袱绉嶏紙鎴栧绉嶏級婧舵恫鐨勮川閲忓姞璧锋潵锛屽幓鎺変笉婧剁墿锛屾瘮濡傛皵浣...
绛旓細鐢辫〃鍙互鐪嬪嚭绗洓娆Cl灏卞弽搴斿畬浜嗭紝鎵浠ョ涓夋鏄伆濂藉弽搴旂殑銆傛牴鎹川閲忓畧鎭掞紝鏍峰搧涓殑纰抽吀閽欎负2.7g鎵浠ヨ川閲忓垎鏁颁负2.7/3=90%2.7g纰抽吀閽欏嵆0.027mol瀹屽叏鍙嶅簲闇瑕丠Cl0.054mol鍗1.971g鎵浠ョ洂閰哥殑璐ㄩ噺鍒嗘暟涓1.971/27=7.3
绛旓細16棰樺鍚?Cu(OH)2+H2SO4=CuSO4+2H2O 鍥犱负浜х墿娌℃湁娌夋穩鍜屾皵浣,鍥犳鍙煡CuSO4璐ㄩ噺鍔犱笂鎵鏈夌殑姘寸殑璐ㄩ噺,灏辩瓑鏂煎師Cu(OH)2鍔犱笂H2SO4婧舵恫鐨勮川閲 鍋囪鍙嶅簲寰屾憾娑蹭负100g,閭ｉ杭CuSO4=16g,杩涜屾眰鍑篊u(OH)2=9.8g,鍥犳鍘烪2SO4婧舵恫璐ㄩ噺涓90.2g 鑰岀～閰告憾璐ㄨ川閲忎篃鑳介氳繃鏂圭▼璁＄畻寰楀埌H2SO4=9.8g,鎵浠2...
绛旓細杩欎釜瑕佹牴鎹綘浠綋鍦鏈鍚庝竴閬撳寲瀛﹂鐨勮瘎鍒嗘爣鍑嗘潵鐪嬩簡锛屽洜涓轰腑鑰冨拰楂樿冪殑闃呭嵎閮芥槸鎸夋楠ょ粰鍒嗙殑锛屼弗鏍兼寜鐓ц瘎鍒嗘爣鍑嗘潵璇勫垎銆傚鏋滆浣犵殑鐩稿鍒嗗瓙璐ㄩ噺璁＄畻閿欎簡鐨勮瘽锛岃繖涓儏鍐垫槸涓嶅お涔愯鐨勶紝鍥犱负浣犲悗闈㈢殑璁＄畻鍏ㄩ兘鏄熀浜庣浉瀵瑰垎瀛愯川閲忚屾潵鐨勶紝濡傛灉鐩稿鍒嗗瓙璐ㄩ噺璁＄畻閿欒鐨勮瘽锛屽鑷村悗闈㈢殑缁撴灉鍙兘閮戒細鏈夐敊璇璁＄畻棰瑙...
