【急】一道高一化学计算题
\u4e00\u9053\u9ad8\u4e00\u5316\u5b66\u8ba1\u7b97\u9898Cu+2H2SO4\u6d53==CuSO4+SO2(\u5347\uff09+2H2OCuO+H2SO4==CuSO4+H2O\u53cd\u5e94\u5269\u4f59\u7684\u786b\u9178\u4e0eNaOH\u53cd\u5e94nNaOH=1mol/L*0.085L=0.085mol\u53c8\u56e0\u4e3a\u4fe9NaOH\u4e0e\u4e00\u4e2a\u786b\u9178\u53cd\u5e94\u3002\u6240\u4ee5n\u5269\u4f59H2SO4=0.0425mol\u7136\u540e\u8fc7\u91cf\u94c1\u7c89\u4e0eCu2+\u7f6e\u6362\u3002Fe+Cu2+==Cu+Fe2+\u94c1\u7684\u6469\u5c14\u8d28\u91cf56g/mol\u3002\u94dc64g/mol\u3002\u6240\u4ee5\u5dee\u6bd4\u6cd5\u3002\u6c42CU\u8d28\u91cf64-56=8 3.6-3.36=0.24\u6240\u4ee5nCU=0.24/8=0.03mol\u3002\u3002\u3002\u3002\u3002\u3002\u3002\u3002\u3002\u3002\u6240\u4ee5\u7269\u8d28\u7684\u91cf\u6d53\u5ea60.03/0.5=0.06mol/Lm\u603bCU=1.92g\u5269\u4f59\u662f\u6c27\u4e3a0.32g\u6240\u4ee5mCUO\uff08\u6c27\u5316\u94dc80g/mol\uff09=0.32/16*80=1.6g\u3002\u3002\u3002\u3002\u3002\u3002\u3002\u3002\u3002\u5171\u6d88\u8017\u786b\u91780.32/16=0.02mol\u6240\u4ee5\u539fCU\u4e3a0.64g\u3002\u3002\u3002\u3002\u3002\u3002\u3002\u3002\u3002\u3002\u3002\u3002\u3002\u3002\u3002\u3002\u3002\u3002\u3002\u3002\u3002\u3002\u3002\u5171\u6d88\u8017\u786b\u91780.64/64*2=0.02mol\u6240\u4ee5\u786b\u9178\u5171\u67090.0825mol\u3002\u3002\u3002\u3002\u3002\u3002\u3002\u3002\u3002\u3002\u3002\u3002\u3002\u3002\u3002\u3002\u3002\u3002\u3002\u6240\u4ee5\u7269\u8d28\u7684\u91cf\u6d53\u5ea6\u4e3a0.0825/0.005=16.5mol/L\u4e0d\u8c22\u3002\u3002\u5c31\u662f\u6709\u70b9\u7d2f\u4eba\u3002\u3002\u3002
\u94dc\u7684\u8d28\u91cf\u4e3a0.2\u514b.\u8bbe\u9541\u4e3ax,\u94dd\u4e3ay,\u5219\uff1a 2*x+3*y=2*1.12/22.4. 24*x+27*y=1.2-0.2=1.0 \u5f97x=0.4,y=0.6.,\u518d\u7531\u7535\u8377\u5b88\u6052\uff1b2Al~3H2\u8f6c\u5316\u4e3a\u6c14\u4f53\u4f53\u79ef\u540e\u5f970.747L
思路分析: 结合氧化锌和锌与盐酸的反应计算。解:设氧化锌物质的量为x,锌的物质的量为y。
Zn + H2SO4 == ZnSO4 + H2↑
65 98 161 2
ZnO + H2SO4 == ZnSO4 + H2O
81 98 161
x+y=0.4×0.5,……①
81x+65y=14.6。……②
联立①②式,得:x=0.1,y=0.1。
0.4mol/L*500ml=0.2mol(含锌离子量)
设原来含锌Xmol
65X+81(0.2-X)=14.6
X=0.1mol
0.2-X=0.1mol
设原混合物中氧化锌和锌的物质的量分别为x,y,
65x+81y=14.6
x+y=0.4*0.5=0.2
x=0.1,y=0.1
设氧化锌和锌的物质的量分别为x,y(mol)
(65.4+16)*x+16*y=14.6
x+y=0.4*0.5
解得x=0.095,y=0.105
都是零点一摩尔
首先求出锌粒子的总物质的量,为0.4*0.5=0.2mol
设锌单质的物质的量为xmol ,氧化锌的物质的量为ymol
那么x+y=0.2
65x+(65+16)y=14.6
解得 x=y=0.1
设ZnO为m,Zn为n,则:
81m+65n=14.6
m+n=0.2
解得:m=0.1,,,n=0.1(mol)
绛旓細锛1锛夊渾搴曠儳鐡讹紝鍧╁煔锛2锛塂锛汣锛汢锛屽垎娑叉紡鏂楋紙3锛夋笚鏋 28锛1锛20%锛4锛2锛22400a/(1000p-17a)绗涓閬撹绠楅鏄川閲忓垎鏁版眰瑙o紝20涔樹互14%+30涔樹互24%姹傚緱姘寲閽犵殑鎬昏川閲忋傚啀闄や互鎬昏川閲50.鍙緱缁撴灉锛岀浜岄棶鏄疌=1000pw/M.绗簩閬撹绠楅鏄厛鎶奱=1000pw/M.鎶妛琛ㄧず鍑烘潵銆倃=姘ㄧ殑璐ㄩ噺/姘ㄧ殑璐ㄩ噺+...
绛旓細锛1锛塀r2 + SO32- + H2O = 2Br- + SO42- + 2H+ 鈶 2H+ + SO32- = H2O + SO2鈫 鈶 (2)5.0绔嬫柟绫 = 5000L 鏈塶(Br2)= 60mol , n(H+)= 10mol 鏍规嵁鏂圭▼寮忊憼锛屾秷鑰梟(SO32-)= 60 mol 涓旂敓鎴 n(H+)= 120mol 鍏辨秷鑰梟(SO32-)= 60mol锛岄渶鍔犲叆V锛圢a2SO3锛= 120L...
绛旓細鎵浠ワ紝璇ョ閰搁摱婧舵恫鐨勮川閲忓垎鏁= (170g/235g) 脳100% = 72.3%銆傜瓟妗堬細72.3 25锛庡垎鏋愶細CO2涓嶯a2O2鍙嶅簲锛屼互鍙奌2O涓嶯a2O2鍙嶅簲鍧囦娇鍥轰綋鐗╄川璐ㄩ噺澧炲姞锛岃繖涓庨缁欏浐浣撹川閲忎粠37.16 g鍙樹负40 g锛屽鍔2.84 g鐩稿叧锛屽彲杩愮敤鍖栧鏂圭▼寮忚繘琛屸滆川閲忓樊閲忔硶鈥濆垪寮璁$畻鍑轰袱绉嶆皵浣撶殑鐗╄川鐨勯噺銆傚彟澶栵紝4.76 g+37....
绛旓細2AL+3H2SO4=AL2(SO4)3+3H2 2 3 3 0.3 x y 8.1g閾濈殑鐗╄川鐨勯噺8.1/27=0.3mol 鐢熸垚姘㈡皵涓簓=0.3*3/2=0.45mol 鏍囧噯鐘跺喌涓嬪弽搴旂敓鎴愮殑h2鐨勪綋绉0.45*22.4=10.08L 纭吀鐨勭墿璐ㄧ殑閲忎负x=0.3*3/2=0.45mol 鎵浠ョ~閰告憾娑茬殑鐗╄川鐨勯噺鐨勬祿搴︿负0.45/200/1000=2.25mol/...
绛旓細锛
绛旓細涓锛 灏70鍏婲a2O2鍜孨a2O鐨勬贩鍚堢墿涓98鍏嬫按鍏呭垎鍙嶅簲锛屾墍寰楃殑NaOH婧舵恫璐ㄩ噺鍒嗘暟涓50%锛岃瘯姹傚師娣峰悎鐗╀腑Na2O2鐨勮川閲忔槸澶氬皯锛熶簩锛庡皢鏍囧噯鐘跺喌涓嬬殑HCl姘斾綋360L婧朵簬1L姘达紝鎵寰桯Cl瀵嗗害涓1.18g/cm3,姹傜洂閰哥殑璐ㄩ噺鍒嗘暟鍜岀墿璐ㄧ殑閲忔祿搴︼紟涓锛庤锛歂a2O2鐨勭墿璐ㄧ殑閲忎负x,Na2O鐨勭墿璐ㄧ殑閲忎负y 78x+62y=70 2(x+y)...
绛旓細绮鹃50閬楂樹竴鍖栧璁$畻棰棰樼洰 2涓洖绛 #鐑# 鑱屽満涓婂彈濮斿眻瑕佷笉瑕佷负鑷繁瑙i噴?鎴戠埍铚昏湏鐐规按0 2013-08-11 路 TA鑾峰緱瓒呰繃1827涓禐 鐭ラ亾灏忔湁寤烘爲绛斾富 鍥炵瓟閲:1140 閲囩撼鐜:25% 甯姪鐨勪汉:261涓 鎴戜篃鍘荤瓟棰樿闂釜浜洪〉 鍏虫敞 灞曞紑鍏ㄩ儴 鏈変竴浠芥皵浣撴牱鍝佺殑璐ㄩ噺鏄14.2 g,浣撶Н鏄4.48 L(鏍囧噯鐘跺喌涓),璇...
绛旓細2NaHCO3=Na2CO3+CO2+H2O 2 1 1 X y 0.05 X=0.1 NaHCO3鐨勮川閲8.4 Na2CO3=19-8.4=10.6(0.1mol)y=0.05 鎬诲叡鍓╀綑Na2CO3鐨勭墿璐ㄧ殑閲=0.1+0.05=0.15 娴撳害=0.15/0.3=0.5
绛旓細V锛圕l2锛=14.2/71*22.4=4.48L H2 + Cl2 = 2HCl 22.4L 22.4L 44.8L 11.2L 4.48L V锛圚Cl锛夋牴鎹鎰忥紝姘㈡皵涓庢隘姘斿簲绛変綋绉弽搴旓紝鎵浠ユ阿姘旀湁鍓╀綑锛屽墿浣6.72L 鐢熸垚姘寲姘㈢殑浣撶Н涓8.96L c锛圚Cl锛=(8.96/22.4)mol/0.25L=1.6mol/L ...
绛旓細棰樼洰搴旇鏄埗鐩愰吀 c(HCl)=1000PW/M=1000*1.19*0.365/36.5=11.9mol/L n1(HCl)=11.9*1.00=11.9(mol/L)V(HCl閫氬叆)=11.9*22.4=266.56L n2(HCl)=1.00*0.25=0.25mol V(娴揌Cl)=0.25/11.9=0.0210L=21.0mL鍙婂彇杩欎箞澶氭祿鐩愰吀绋閲婅嚦250mL ...