标况下224毫升二氧化碳气体通入150毫升0.1莫尔每升的氢氧化钠溶液中的离子方程式
n(NaOH)=0.1mol/L×0.15L=0.015moln(CO2)=0.224/22.4=0.01mol
因为CO2+2NaOH=Na2CO3 + H2O
1mol 2mol
x 0.015
x=0.0075mol
二氧化碳过量
CO2+OH- =HCO3-
绛旓細n(NaOH)=0.1mol/L脳0.15L=0.015mol n(CO2)=0.224/22.4=0.01mol 鍥犱负CO2+2NaOH=Na2CO3 + H2O 1mol 2mol x 0.015 x=0.0075mol 浜屾哀鍖栫⒊杩囬噺 CO2+OH- =HCO3-
绛旓細n(HCl)=224/22.4=10mol m(HCl)=365g HCl%=36.5 c(HCl)=1000*10/(1000/1.18)=11.8mol/L c=11.8*100/1180=1mol/L
绛旓細n(H2) = (224/1000)/22.4 = 0.01 mol 2Na + 2H2O = 2NaOH + H2 n(Na) = 2*n(H2) = 0.02 mol m(Na) = 23*0.02 = 0.46g (1) 閲戝睘閽犳牱鍝佷腑鍚噾灞為挔0.46鍏 m(Na2O) = 0.77 - 0.46 =0.31 g n(Na2O) = 0.31/62 = 0.005 mol Na2O + H2O = 2NaOH n(N...
绛旓細n(CO2)=0.224L/22.4L/mol=0.01mol Na2CO3+2HCl===2NaCl+H2O+CO2鈫 1 1 x 0.01mol x=0.01mol m(Na2CO3)=O.O1mol*106g/mol=1.06g 娉細浠ヤ笂鍋氭硶鍓嶆彁鏄繀椤诲湪鏍囧喌涓锛屽惁鍒欐棤娉曠洿鎺ラ氳繃姘斾綋浣撶Н姹傛皵浣撶殑鐗╄川鐨勯噺锛涘苟涓旂⒊閰搁挔涓嶈繃閲 ...
绛旓細c=m/V=10/1/1.18 杩欎釜浣撶Н鐨勭畻娉曟垜鐪嬩笉鎳 鎶婂崟浣嶅甫涓婂氨鑳芥噦浜嗭紝鍖栧璁＄畻涓紝鑳界敤鍗曚綅鍒ゆ柇姝ｈ V=m/瀵嗗害=锛635g+365g)/(1.18g/mL*1000mL/L锛=1/1.18L 鎶婂瘑搴﹀崟浣嶄腑鐨刴L杞寲涓轰簡L銆傛浣撶Н涓烘憾瑙Cl鍚庢憾娑茬殑浣撶Н c=n/V=10mol/(1/1.18L)=11.8mol/L ...
绛旓細浜х敓鐨姘斾綋閲忎负(224/1000)/22.4=0.01mol 鍙互寰楀埌鍙嶅簲Fe鐨勭墿璐ㄧ殑閲忎负0.01mol锛屾秷鑰楃殑鐩愰吀閲忎负0.02mol锛屾墍娑堣楃殑鐩愰吀婧舵恫涓(0.02/2)*1000=10ml 鍔犲叆KSCN鍚庢棤绾㈣壊锛岃鏄庢棤Fe3+鍗矲e3+琚獸e瀹屽叏杩樺師銆傚張鐩愰吀姝ｅソ鍏ㄥ弽搴斿彲鐭ワ紝鍓╀綑鐨勭洂閰告濂戒笌Fe2O3瀹屽叏鍙嶅簲 6HCl+ Fe2O3==2FeCl3+3H2O 鐩愰吀...
绛旓細鍥犱负涓ゆ鎵寰CO2 浣撶Н涓嶅悓锛岃鏄15mL Na2CO3鐩稿浜25mL 鐩愰吀鏉ヨ 杩囧墿銆傗憼閲忓彇25.00mL鐢叉憾娑诧紝鍚戝叾涓紦鎱㈡淮鍔犱箼婧舵恫15.00mL,鍏辨敹闆嗗埌浜屾哀鍖栫⒊224mL銆俷(CO2) = 224mL/22.4mol/L = 0.01 mol 2HCl + Na2CO3 =2NaCl + H2O + CO2鈫 HCl瀹屽叏鍙嶅簲锛汵a2CO3鍓╀綑銆俷(HCl) = 2n(CO2) = 2*0....
绛旓細鍦ㄦ爣鍑嗙姸鍐典笅锛屼换浣1鎽╁皵姘斾綋鎵鍗犱綋绉兘鏄22.4鍗囥傚彲瑙侊紝鏍囧喌涓224姣崌鐨勬爱姘旀湁0.01鎽╁皵銆傞樋浼忎冀寰风綏甯告暟鏄 6.02*10^23 涓/鎽╁皵锛岄偅涔0.01鎽╁皵鐨勬爱姘斾腑鍒嗗瓙鏁版槸 6.02*10^21涓紝瀵瑰簲鐨勫師瀛愭暟鏄 2*6.02*10^21=1.204*10^22涓紝姘厓绱犳槸7鍙峰厓绱狅紝鍗虫瘡涓爱鍘熷瓙鏈7涓數瀛愶紝鎵浠ユ湰棰樹腑鎬荤數瀛...
绛旓細Fe2O3+3CO=2Fe+3CO2 160 84 1.6 0.84 0.84g涓0.01mol锛鏍囧喌涓22.4*0.01*1000=224姣崌
绛旓細姘寲姘姘斾綋10mol婧朵簬姘达紝婧舵恫鐨勪綋绉笉鏄635ml锛岃屾槸锛氾紙365g+635g)/锛1.18g/銕³锛=847銕³涓轰粈涔堜綋绉鍔犱簡(847銕³-635銕³=212銕³锛熸槸鍥犱负鏈 10mol鐨勬隘鍖栨阿姘斾綋婧朵簬635銕³姘翠腑锛屽鍔犱簡涓浜涗綋绉傛皵浣撳垎瀛愭槸鏈変綋绉殑,鍦ㄨ繖閲屾槸浣撶Н澧炲姞锛屾墍浠ヤ笉鑳界敤 10mol...
