一道高一的化学计算题
\u4e00\u9053\u5316\u5b66\u9ad8\u4e00\u8ba1\u7b97\u9898Mg\u548cNaOH\u4e0d\u53cd\u5e94
\u800c2Al+2NaOH+2H2O=2NaAlO2+3H2\u2191
\u90a3\u4e48Mg\u7684\u8d28\u91cf\u662fcg
Al\u7684\u8d28\u91cf\u662fa-cg
\u800c\u751f\u6210\u7684\u6c22\u6c14\u4f53\u79ef\u662fbmL=b/1000L
\u90a3\u4e48\u7269\u8d28\u7684\u91cf\u662f(b/1000)/22.4=b/22400mol
\u8bbeAl\u7684\u76f8\u5bf9\u539f\u5b50\u91cf\u662fM
\u5219Al\u7684\u7269\u8d28\u7684\u91cf\u662f\uff08a-c\uff09/M
\u6839\u636e\u6bd4\u4f8b\u5f0f\u77e5\u9053\u4e86Al\u7684\u7269\u8d28\u7684\u91cf\u548cH2\u7684\u7269\u8d28\u7684\u91cf\u7684\u5173\u7cfb\u662f
2:3=\uff08a-c\uff09/M\uff1ab/22400
\u89e3\u5f97M=33600\uff08a-c\uff09/b
2Na+2H20=2NaOH+H2
n(NaOH)=n(Na)=2n(H2)=6.9g/23g/mol=0.3mol
2NaOH+Si+H2O=Na2SiO3+2H2
n(NaOH):n(Si)=2:1
\u7531\u4e0a\u53ef\u77e5\u8be5\u53cd\u5e94\u6070\u597d\u5b8c\u5168\u53cd\u5e94n(NaOH)=2n(Si)=2(Na2SiO3)=n(H2)=0..3mol
V(H2)=(0.3/2+0.3)mol*Vm=10.08L
\u6700\u540e\u6eb6\u6db2\u4e2d\u53ea\u5269\u4e0bNa2SiO3
c(Na2SiO3)=0.3/2mol/0.1L=1.5mol/L
\u5316\u5b66\u9898\u56de\u7b54\u771f\u9ebb\u70e6\u554a
\u4e00\u5b9a\u8981\u7ed9\u5206\u554a
2 1 1
X y 0.05 X=0.1 NaHCO3的质量8.4 Na2CO3=19-8.4=10.6(0.1mol)
y=0.05
总共剩余Na2CO3的物质的量=0.1+0.05=0.15 浓度=0.15/0.3=0.5
绛旓細鎬濊矾鍒嗘瀽锛 缁撳悎姘у寲閿屽拰閿屼笌鐩愰吀鐨勫弽搴璁＄畻銆傝В锛氳姘у寲閿岀墿璐ㄧ殑閲忎负x锛岄攲鐨勭墿璐ㄧ殑閲忎负y銆俍n + H2SO4 == ZnSO4 + H2鈫 65 98 161 2 ZnO + H2SO4 == ZnSO4 + H2O 81 98 161 x+y=0.4脳0.5锛屸︹︹憼 81x+65y=14.6銆傗︹︹憽 鑱旂珛鈶犫憽寮忥紝寰楋細x=0.1锛寉=0....
绛旓細1.V=22.4*n=22.4*(m/M)=22.4*(2.2/44)=1.12L 2.M=m/n=m/(V/22.4)=1.92/(672/1000*22.4)=64 3.Zn + 2HCl = ZnCl + H2 1 x=1 V=22.4*nH2=22.4*nZn=22.4*m/M=22.4*26/65=8.96L 4.nKCl:nNaCl=nK:nNa=3:1 mKCl:nNaCl=MKCl*nKCl:MNaCl*nNaCl...
绛旓細绮鹃50閬撻珮涓鍖栧璁＄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)...
绛旓細绛旀锛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+37.16 g锛40 g=1.92 g锛1.92 g鍗虫槸鐢熸垚鐨勬哀姘旂殑璐ㄩ噺锛屽悓鏍峰彲浠...
绛旓細80 0.62g z 62/0.62=80/z z=0.8g 鍙嶅簲鐢熸垚NaOH婧惰川鐨勮川閲忔槸8+0.8=8.8g 鍙嶅簲鍚庢憾娑茬殑璐ㄩ噺鏄5.22+94.98-0.2=100g NaOH鐨勮川閲忓垎鏁版槸8.8/100*100%=8.8 绛旓細锛1锛夊叾涓哀鍖栭挔鐨勮川閲忔槸0.62g銆傦紙2锛夊弽搴斿悗鎵寰桸aOH婧舵恫鐨勮川閲忓垎鏁版槸8.8 ...
绛旓細NaOH鍦ㄦ娓╁害涓嬬殑婧惰В搴︽槸22g 閭ｄ箞楗卞拰NaOH婧舵恫婧惰剛鐨勮川閲忓垎鏁颁负22/122*100%=18 璁惧姞鍏ラ挔鐨勮川閲忎负x 2Na+2H2O=2NaOH+H2鈫 46---80---2 x---m1---m2 瑙ｅ緱m1=40x/23锛宮2=x/23 閭ｄ箞婧舵恫鐨勮川閲忎负x+100-x/23=锛100+22x/23锛塯 NaOH鐨勮川閲忎负40x/23 鍥犱负婧舵恫鏄ケ鍜屾憾娑 鎵浠ユ憾鑴傜殑璐ㄩ噺...
绛旓細涔熷氨鐭ラ亾鍓╀綑鍥轰綋蹇呯劧鍏ㄩ儴鏄疦a2CO3銆傝鍘熸贩鍚堢墿涓璑aHCO3涓簒mol锛孨a2CO3涓簓mol锛屾牴鎹師娣峰悎鐗╄川閲忓緱鏂圭▼锛84x+106y=100锛屾牴鎹甆a瀹堟亽寰楁柟绋嬶細(x+2y+20/40)梅2脳106=95.5 瑙ｆ柟绋嬬粍锛歺=1锛寉=8/53銆傚師娣峰悎鐗╀腑纰抽吀閽犺川閲忎负106g/mol脳8/53mol=16g锛岃川閲忓垎鏁颁负16g梅100g脳100%=16%銆
绛旓細瑙ｆ硶涓:(1)2NaI+Cl2=2NaCl+I2 ---2*150---2*58.5---254 ---m(NaI)---m1(NaCl)--6.35g 鍒楁瘮渚嬪紡瑙ｅ緱锛歮(NaI)=7.5g,m1(NaCl)=2.925g NaCl+AgNO3=AgCl鈫+NaNO3 ---58.5---143.5 ---m鎬(NaCl)--28.7g 鍒楁瘮渚嬪紡瑙ｅ緱锛歮鎬(NaCl)=11.7g 鎵浠ワ紝鍘熸贩鍚堢墿涓璑aCl鐨勮川...
绛旓細璐ㄩ噺鍑忓皯 2NaHCO3=Na2CO3+CO2+H2O 168g 62g x 0.31g 瑙ｅ緱锛歺=0.84g 璐ㄩ噺鍒嗘暟锛0.84/1.9*100%=44.2
