上海化学高考题
\u4e0a\u6d77\u5316\u5b66\u9ad8\u8003\u9898[\u7b54\u6848]\uff1a\uff081\uff090.012mol\u3002(2\uff093.8\u3002(3\uff09n(OH\uff0d)\u2236n(CO32\uff0d)\uff1d16\u22361\u3002 (4\uff09Al2Mg6(OH)16CO3•4H2O\u3002
[\u89e3\u6790]\uff1a(1)2\u7247\u5c0f\u82cf\u6253\u6240\u542bNaHCO3\u7684\u8d28\u91cf\u4e3a1g\uff0c\u5176\u7269\u8d28\u7684\u91cf\u4e3a1g84g/mol =0.012mol\uff0c
\u7531HCO3\uff0d\uff0bH+===CO2\u2191\uff0bH2O\uff0c\u53ef\u77e5n(H\uff0b)=0.012mol\u3002
(2)6\u7247\u5c0f\u82cf\u6253\u7684\u7269\u8d28\u7684\u91cf\u4e3a\uff1a3/84==0.036mol\uff0c\u5373\u4e2d\u548c\u7684H\uff0b\u4e3a0.036mol\uff0c\u800c\u6bcf\u7247\u7684Al(OH)3\u7684\u7269\u8d28\u7684\u91cf\u4e3a0.245/78=0.0031mol\uff0c\u7531Al(OH)3\uff0b3H\uff0b ===Al3+\uff0b3H2O\uff0c\u6240\u4ee5Al(OH)3\u7684\u7247\u6570\u4e3a\uff1a0.012/0.0031=3.8\u7247\u3002
(3)\u2460\u78b1\u5f0f\u76d0\u4e2d\u52a0\u5165HCl\uff0c\u9996\u5148\u662f\u78b1\u5f0f\u76d0\u4e2d\u7684OH-\u548cCO32-\u4e0eH+\uff0842.5mL\uff09\u53cd\u5e94\u751f\u6210H2O\u548cHCO3-\uff0c,\u7136\u540eHCO3-\u7ee7\u7eed\u4e0e\u76d0\u9178\u53cd\u5e94\uff0c\u6d88\u801745.0mL\u65f6\u6b63\u597d\u53cd\u5e94\u5b8c\u5168\u3002
\u6240\u4ee5n(HCO3\uff0d)\uff1d2.0mol•L\uff0d1\u00d7\uff080.045L-0.0425L\uff09\uff1d0.005mol\uff0c\u5373n(CO32-)=0.005mol\uff0cH\uff0b\u4e0eCO32-\u3001OH-\uff0d\u53cd\u5e94\u7684H\uff0b\u7684\u603b\u7269\u8d28\u7684\u91cf\u4e3a\uff1a2.0mol•L\uff0d1\u00d70.0425L\uff1d0.085mol\uff0c\u6240\u4ee5n(OH-)=0.08mol\uff0c\u8be5\u78b1\u5f0f\u76d0\u6837\u54c1\u4e2d\u6c22\u6c27\u6839\u4e0e\u78b3\u9178\u6839\u7684\u7269\u8d28\u7684\u91cf\u4e4b\u6bd4\uff1a16\u22361\u3002
\u2461\u80fd\u4e0e\u8fc7\u91cf\u7684NaOH\u6eb6\u6db2\u53cd\u5e94\u4ea7\u751f\u6c89\u6dc0\u7684\u53ea\u6709Mg2\uff0b\uff0c\u6240\u4ee5n[Mg(OH)2]\uff1d1.74/58=0.03mol\uff0c\u82e5\u78b1\u5f0f\u76d0\u4e2d\u4e0d\u542b\u7ed3\u6676\u6c34\uff0c\u5219\u6c22\u5143\u7d20\uff08OH-\uff09\u7684\u8d28\u91cf\u5206\u6570\u4e3a\uff1a0.08mol\u00d71g/mol\uff1d0.08g\uff0c\u6c22\u5143\u7d20\u7684\u8d28\u91cf\u5206\u6570\u4e3a\uff1a0.08/3.01=0.027\uff1c0.04\uff0c\u8bf4\u660e\u78b1\u5f0f\u76d0\u4e2d\u542b\u6709\u7ed3\u6676\u6c34\uff0c\u6839\u636e\u9898\u610f\u6709\uff1a
m(Al)\uff0bm(H2O)\uff0b0.03mol\u00d724g/mol\uff08Mg2+\uff09\uff0b0.005mol\u00d760g/mol\uff08CO32-\uff09\uff0b0.08mol\u00d717g/mol\uff08OH-\uff09=3.01g\uff0c[2/18*m\uff08H2O\uff09+0.08*1]/3\u300201=0.04\uff0c\u6240\u4ee5m(H2O)=0.36g\uff0cn(H2O)=002mol\uff0cm(Al)=0.27g\uff0cn(Al)=001mol\uff0c\u6240\u4ee5n(Al3\uff0b)\u2236n(Mg2\uff0b)\u2236n(OH\uff0d)\u2236n(CO32\uff0d)\u2236n(H2O)
\uff1d0.01mol\u22360.03mol\u22360.08mol\u22360.005mol\u22360.02mol\uff1d2\u22366\u223616\u22361\u22364\uff0c\u5373\u78b1\u5f0f\u76d0\u7684\u5316\u5b66\u5f0f\u4e3a\uff1aAl2Mg6(OH)16CO3•4H2O\u3002
\u5e0c\u671b\u5bf9\u4f60\u80fd\u6709\u6240\u5e2e\u52a9\u3002
\u6c14\u4f53\u4e0d\u8db3\u91cf\u65f6\uff08\u5373\u9541\u8fc7\u91cf\uff09\u4e5f\u662f\u53ef\u80fd\u7684\u3002m>2.4g .
\u9541\u6761\u5728\u53ea\u542b\u6709CO2\u65f6\uff0c\u6700\u7ec84.6g.
\u9541\u6761\u5728\u53ea\u542b\u6709O2\u65f6\uff0c\u6700\u7ec84.0g.
\u8303\u56f42.4<m<4.6
\u4e0d\u80fd\u9009\u6781\u503c\u3002\u9009D
[解析]：(1)2片小苏打所含NaHCO3的质量为1g，其物质的量为1g84g/mol =0.012mol，
由HCO3－＋H+===CO2↑＋H2O，可知n(H＋)=0.012mol。
(2)6片小苏打的物质的量为：3/84==0.036mol，即中和的H＋为0.036mol，而每片的Al(OH)3的物质的量为0.245/78=0.0031mol，由Al(OH)3＋3H＋ ===Al3+＋3H2O，所以Al(OH)3的片数为：0.012/0.0031=3.8片。
(3)①碱式盐中加入HCl，首先是碱式盐中的OH-和CO32-与H+（42.5mL）反应生成H2O和HCO3-，,然后HCO3-继续与盐酸反应，消耗45.0mL时正好反应完全。
所以n(HCO3－)＝2.0mol•L－1×（0.045L-0.0425L）＝0.005mol，即n(CO32-)=0.005mol，H＋与CO32-、OH-－反应的H＋的总物质的量为：2.0mol•L－1×0.0425L＝0.085mol，所以n(OH-)=0.08mol，该碱式盐样品中氢氧根与碳酸根的物质的量之比：16∶1。
②能与过量的NaOH溶液反应产生沉淀的只有Mg2＋，所以n[Mg(OH)2]＝1.74/58=0.03mol，若碱式盐中不含结晶水，则氢元素（OH-）的质量分数为：0.08mol×1g/mol＝0.08g，氢元素的质量分数为：0.08/3.01=0.027＜0.04，说明碱式盐中含有结晶水，根据题意有：
m(Al)＋m(H2O)＋0.03mol×24g/mol（Mg2+）＋0.005mol×60g/mol（CO32-）＋0.08mol×17g/mol（OH-）=3.01g，[2/18*m（H2O）+0.08*1]/3。01=0.04，所以m(H2O)=0.36g，n(H2O)=002mol，m(Al)=0.27g，n(Al)=001mol，所以n(Al3＋)∶n(Mg2＋)∶n(OH－)∶n(CO32－)∶n(H2O)
＝0.01mol∶0.03mol∶0.08mol∶0.005mol∶0.02mol＝2∶6∶16∶1∶4，即碱式盐的化学式为：Al2Mg6(OH)16CO3•4H2O。
①滴加盐酸的过程中分别发生了3步反应:第一: H+ + OH- ==== H2O第二: H+ + CO32- ==== HCO3-第三: H+ + HCO3- ==== CO2 + H2O结合已知条件可知,反应三消耗的盐酸的体积为2.5mL,那第二过程消耗的盐酸体积也应该为2.5mL因此n(OH-):n(CO32-)=40:2.5=16:1②经分析可知所得1.74g沉淀为Mg(OH)2,则n(Mg2+)=n(Mg(OH)2)=1.74/58=0.03mol又n(OH-)=40mL×2mol/L=0.08moln(CO32-)=0.08/16=0.005mol利用电荷守恒可列等式: 2n(Mg2+)+3n(Al3+)=2n(CO32-)+n(H+)因此求得n(Al3+)=0.01mol结合各微粒物质的量关系可知化学式为: Mg6Al2(OH)16CO3最后利用氢元素质量为m(H)=3.01×0.04=0.1204gn(H)=m(H)/1=0.12mol>n(OH-)因此该化合物中还含有其他含H元素的微粒,自然会联想到结晶水.有m(H2O)=3.01-24×0.03-27×0.01-0.08×17-0.005×60=0.36则n(H2O)=0.02mol最终确定该碱式盐化学式为:Mg6Al2(OH)16CO3·4H2O
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