【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)
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