欲使滴定时消耗0.1mol/LKMnO4溶液20-25ml,需称取基准试剂Na2C2O4多少克？ 欲使滴定时消耗0.1mol/Lhcl20-25ml 问应称取...
\u6b32\u4f7f\u6ef4\u5b9a\u65f6\u6d88\u80170.10mol/LHCI\u6eb6\u6db220-25mL\uff0c\u95ee\u5e94\u53d6\u57fa\u51c6\u8bd5\u5242Na2CO3\u591a\u5c11\u514b\uff1f\u6c42\u65510.1X20X106/2X1000\u52300.1X25X106/2X1000
\u53730.106\u514b\u52300.1325\u514b\u4e4b\u95f4
0.2g\u5de6\u53f3\u3002
按照反应式可以计算出来的，2KMnO4+5Na2C2O4+16H+=2K++2Mn2++10CO2+8H2O+10Na+
草酸钠的摩尔数为：0.002-
0.0025mol，上下限乘以草酸钠的摩尔质量134，得到了0.268-0.335克
绛旓細鑽夐吀閽犺川閲忕害涓0.67g锝0.8375g 鍙嶅簲鏂圭▼寮:5Na₂C₂O₄锝2KMnO₄锛堥吀鎬ф潯浠讹級= 2Mn2+ 锛10CO₂锛8H2O锛岄渶瑕婊村畾楂橀敯閰搁捑閲忎负锛0.002锝0.0025mol锛屾牴鎹瀛愬弽搴旀柟绋嬪紡鍙煡闇瑕佽崏閰搁挔锛0.005锝0.00625mol锛屾墍浠ラ渶瑕佽崏閰搁挔璐ㄩ噺锛134g/mol脳0.005锝0.00625锛0...
绛旓細0.0025mol锛屼笂涓嬮檺涔樹互鑽夐吀閽犵殑鎽╁皵璐ㄩ噺134锛屽緱鍒颁簡0.268-0.335鍏
绛旓細瑕佹兂娑堣鐩愰吀鐨勪綋绉湪20鈥25姣崌涔嬮棿锛岀О鍙栫⒊閰搁挔鐨勮川閲忚寖鍥存槸0.1056鈥0.132g銆
绛旓細KHP鏄竴绉嶅父鐢ㄧ殑閰稿紡鐩愶紝鍏跺寲瀛﹀紡涓篕HC8H4O4銆傝姹傚湪婊村畾涓秷鑰0.1 mol/L NaOH 婧舵恫 25-30 ml锛屽彲浠ラ氳繃KHP鐨勬懇灏旇川閲忓拰婊村畾鍙嶅簲鐨勫寲瀛︽柟绋嬪紡鏉ヨ绠楁墍闇鐨勮川閲忋傞鍏堬紝鍖栧鏂圭▼寮忎负锛欿HC8H4O4 + NaOH 鈫 KNaC8H4O4 + H2O 璇ユ柟绋嬪紡琛ㄦ槑锛屽湪婊村畾杩囩▼涓紝姣1 mol鐨 KHC8H4O4 浼氫笌1 mol...
绛旓細濡傛灉娑堣0.1mol/L鏍囧噯婧舵恫鐨勪綋绉秴杩囦簡0.2ml锛岄偅涔堝彲鑳藉瓨鍦ㄤ互涓嬮棶棰橈細1. 楗叉枡涓矖铔嬬櫧鍚噺杈冮珮锛屽鑷婊村畾缁堢偣鎻愬墠鍒版潵锛屼粠鑰屾秷鑰楃殑鏍囧噯婧舵恫浣撶Н鍋忓皬銆2. 鍦ㄦ搷浣杩囩▼涓锛屽彲鑳藉瓨鍦ㄨ宸垨鑰呬笉褰撴搷浣滐紝瀵艰嚧娑堣楃殑鏍囧噯婧舵恫浣撶Н鍋忓皬銆3. 楗叉枡涓彲鑳藉惈鏈夊叾浠栬兘澶熶笌鏍囧噯婧舵恫鍙嶅簲鐨勭墿璐紝瀵艰嚧娑堣楃殑鏍囧噯婧舵恫浣撶Н鍋...
绛旓細c(HCl)锛漨/0.05299脳V=0.1124/(0.05299*20.02)=0.1060(mol/L)寮涓锛歝(HCl)鈹鈹鐩愰吀婧舵恫鐨勭墿璐ㄧ殑閲忔祿搴︼紱m鈹鈹绉板彇鏃犳按纰抽吀閽犺川閲忥紝g锛沄鈹鈹婊村畾鐢ㄥ幓鐩愰吀婧舵恫瀹為檯浣撶Н锛宮L锛0.05299鈹鈹涓1.00mL鐩愰吀婧舵恫銆攃(HCl)=1.000mol銆曠浉褰撶殑浠ュ厠琛ㄧず鐨勬棤姘寸⒊閰搁挔鐨勮川閲忋
绛旓細0.1mol纭濋吀閾剁┖鐧婊村畾娑堣0.1~0.2姣崌銆傛憾娑涓CrO42绂诲瓙绾︾瓑浜0.01mol婧舵恫涓湭鐢熸垚Ag2CrO4鐨凙g绂诲瓙绛変簬5mol鐢熸垚鐨凙g2CrO4涓瑼g绂诲瓙绛変簬2涔10-5mol鎬荤殑婧舵恫涓瑼g绂诲瓙绛変簬3.4涔10-5mol纭濋吀閾跺皯閲忥紝浣撶Н鍙樺寲蹇界暐锛岀瓑浜0.085ml绾0.1ml銆
绛旓細瑙ｏ細(1)NaOH+KHC8H4O4=KNaC8H4O4+H2O婊村畾鏃舵秷鑰0.1molL1NaOH婧舵恫浣撶Н涓20mL鎵闇绉板彇鐨凨HC8H4O4閲忎负锛歮1=0.1molL120mL103204gmol1=0.4g婊村畾鏃舵秷鑰0.1molL1NaOH婧舵恫浣撶Н涓30mL鎵闇绉板彇鐨凨HC8H4O4閲忎负锛歮2=0.1molL130mL103204gmol1=0.6g鍥犳锛屽簲绉板彇KHC8H4O4鍩哄噯鐗╄川0.4~0....
绛旓細鎵闇Na2CO3鐨勬懇灏旇川閲忥細500 ml * 0.1000 mol/l = 0.0500mol Na2CO3鐨勬懇灏旇川閲忥細23 * 2 + 12 + 16 *3锛106 g/mol 鎵浠ュ彇0.0500 mol * 106g/mol = 5.3g
绛旓細0.1X20X106/2X1000鍒0.1X25X106/2X1000 鍗0.106鍏嬪埌0.1325鍏嬩箣闂
