一道高一化学较难题，求解！
\u4e00\u9053\u9ad8\u4e00\u5316\u5b66\u96be\u9898\u53c2\u52a0\u53cd\u5e94\u7684AgNO3\u7684\u7269\u8d28\u7684\u91cf\u662f
(15.46-4.66)/108 = 0.1(mol)
BrClx \u2014\u2014\u2014 (x+1)AgNO3
80+35.5x x+1
4.66g 0.1 mol
\u5217\u5f0f: (80+35.5x):4.66 = (x+1):0.1
\u89e3\u5f97 x = 3
\u8fd9\u6837\u53ef\u4ee5\u4e86\u5417 \u6211\u53ef\u662f\u7b2c\u4e00\u4e2a\u554a~ ..\u7ed9\u6211\u5206\u5427....
\u5229\u7528\u4e09\u6bb5\u5f0f\u5f88\u5bb9\u6613\u505a\u51fa\u6765\u8fd9\u7c7b\u9898
\u5148\u5199\u51fa\u65b9\u7a0b\u5f0f 2SO2 + O2 ====== 2SO3
\u8d77\u59cb\u65f6\u7269\u8d28\u7684\u91cf 0.70 0.40 0
\u53d8\u5316\u7684\u7269\u8d28\u7684\u91cf 2x x 2x
4s\u672b\u65f6\u7269\u8d28\u7684\u91cf 0.30 0.4-x 2x
\u5219\u67090.7-2x=0.3 \u5f97x=0.2mol
\u6240\u4ee5\u0394n(O2)=0.2mol
v(O2)=\u0394n(O2)/(V*t)=0.2/\uff084*4\uff09=0.0125mol*L^-1*s^-1
2. 3 Cu + 8 HNO3====3Cu(NO3)2 + 2NO +4H2O
3.2(NH4)2CO3+ H2SO4==(NH4)2SO4+ 2 NH4HCO3
绛旓細1 930kg*1000/(44g/mol) *22.4 = 473455L 930kg*1000/(44g/mol) *NA = 21136NA 735kg*1000/(32g/mol) *22.4 = 514500L 2 0.3 mol鐨勬哀姘 0.3*32=9.6g 0.2 mol鐨勮嚟姘(O3) 0.2*48=9.6g 鍥犳璐ㄩ噺姣斾负1:1锛屽垎瀛愭暟姣斾负涔熶负0.3:0.2=3:2 鍘熷瓙涓暟姣斾负0.3...
绛旓細2Na2O2+2CO2=2Na2CO3+O2 姣2molCO2鍙嶅簲锛屽浐浣撳閲88-32=56g 鐒跺悗閫氳繃Na2O2锛屽閲嶄簡7.0g 鎵浠O2涓7/56*2=0.25mol 璁綜2H6 xmol锛孋3H8 ymol 鏈2x+3y=0.25锛6x+8y=0.35*2 x=y=0.5 浣撶Н姣斿氨鏄懇灏旀瘮=1锛1
绛旓細2.璁鹃渶瑕乂 ml鐨勬祿鐩愰吀锛 V*1.19*0.37/36.5=HCl鐨勬懇灏旈噺=2L*0.2锛屾眰寰梀鍗冲彲銆
绛旓細瑙ｏ細涔欏師瀛愭湁涓変釜鐢靛瓙灞傦紝绗竴灞備笌鏈澶栧眰鐢靛瓙鏁扮浉绛夆啋绗竴灞傜數瀛愭暟涓2锛屾墍浠ヤ箼鍘熷瓙搴忔暟涓12鈥擬g 鐢插師瀛愮殑鏍稿鐢靛瓙鏁版瘮涔欏師瀛愭牳澶栫數瀛愭暟灏1鈫掔敳鈥11鍙峰厓绱犫擭a 涓欏師瀛愮殑鏈澶栧眰鐢靛瓙鏁扮殑娆″灞傜數瀛愭暟鐨2鍊嶏細鏈澶栧眰鐢靛瓙鏁颁笉瓒呰繃8锛屾澶栧眰涓2銆8銆18涓夌锛屼絾涓虹煭鍛ㄦ湡鍏冪礌锛屼笝鍘熷瓙搴忔暟涓6鈥擟 涓佸師...
绛旓細4 8 0 X 3X 2X 4-X 8-3X 2X (2X)/(4-X+8-3X+2X)=0.2 瑙ｅ緱X=1 鎵浠ュ墿浣欐阿姘斾负5mol銆傛荤墿璐ㄧ殑閲10mol.姘㈡皵鐧惧垎姣斾负50 锛2锛夊弽搴斿墠浣撶Н鐗╄川鐨勯噺12.鍙嶅簲鍚10.浣撶Н姣1姣1锛屽帇寮烘瘮5姣6 锛3锛塜/4=0.25锛堟爱姘旇浆鍖栫巼锛 3X/8=0.375锛堟阿姘旇浆鍖栫巼锛夛紙4锛...
绛旓細AI2<SO>3鐗╄川鐨勯噺鏄342 MgSO4 鐗╄川鐨勯噺鏄120 娣峰悎鐗╀负2mol 涓擜L2銆奡O4銆:MgSO4涓2锛3 鍒橝L2銆奡O4銆嬫湁0.8mol MgSO4鏈1.2mol 鎵浠 MgSO4 涓144鍏
绛旓細A:H B:C C : N D: O E : F F :S 2. 3 Cu + 8 HNO3===3Cu(NO3)2 + 2NO +4H2O 3.2(NH4)2CO3+ H2SO4==(NH4)2SO4+ 2 NH4HCO3
绛旓細c(HCl )=n(HCl)/V=0.00118a/(a*10-3)=1.18 mol/L 鎵浠 鈶犳槸1.18 mol/L 鈶℃湁C=n/V 鍙互绠楀嚭锛岄渶瑕佺殑娴撶洂閰革紱鐜荤拑妫掞紝瀹归噺鐡讹紝婊寸锛岀儳鏉 鈶 B (3)n(Na2CO3)=0.795/106=0.0075 mol Na2CO3+2HCl =2 NaCl +H2O +CO2鈫 1 2 0.0075mol x 鎵浠=0.0075mol ...
绛旓細O2鐨勮川閲=33.2g-28.4g =4.8g 璁惧弽搴旂敓鎴愮殑KCl璐ㄩ噺涓篨,鍒嗚В鐨凨ClO3璐ㄩ噺涓簓 2KClO3==2KCl+3O2 245^^^149^^^96 y^^^X^^^4.8 X=7.45 y=12.25 (1)KCl+AgNO3==AgCl+KNO3 ^^^74.5^^^143.5 ^^^7.45^^^z z=14.35g (2)MnO2+4HCL(娴)鈫扢nCl2+Cl2+2H2O ^^^87^^^1mo...
绛旓細Mg---Mg锛圢O3)2---2NaOH Mg澶卞幓2涓數瀛愶紝娑堣2涓狽aOH Cu---Cu(OH)2 Mg---Mg(OH)2 绠楀嚭OH-鐨勮川閲忓姞涓婇摐鍜岄晛鐨勫悎閲4銆6g灏辨槸娌夋穩鐨勮川閲 OH-鐨勭墿璐ㄧ殑閲=NaOH鐨勭墿璐ㄧ殑閲=杞Щ鐢靛瓙鐗╄川鐨勯噺 4480ml鐨凬O2姘斾綋涓0.2鎽╁皵 336ml鐨凬2O40.015鎽╁皵 杞Щ鐢靛瓙鐨勭墿璐ㄧ殑閲=0.2*1+0.015*...
