解:(1)令溶液中Al
3+的物質(zhì)的量為a mol,由電荷守恒可知Mg
2+的物質(zhì)的量為
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,溶液中離子總的物質(zhì)的量為三種離子之和,也等于
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mol,故
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+amol+0.7mol=
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mol,解得a=
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,由極限法可知0<a<
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,故0<
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<
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,解得0<x<0.25,故答案為:0<x<0.25;
(2)溶液中的陽離子恰好完全沉淀,此時溶液中溶質(zhì)為KCl,根據(jù)守恒可知n(KOH)=n(KCl)=n(Cl
-)=0.7mol,故KOH溶液是體積為
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=0.0875L=87.5mL,故答案為:87.5;
(3)當(dāng)100 mL KOH剛好完全消耗,且Al
3+全部轉(zhuǎn)化為AlO
2-時,此時溶液中溶質(zhì)為KCl、KAlO
2,根據(jù)守恒可知n(K
+)=n(Cl
-)+n(AlO
2-),故n(AlO
2-)=0.1L×8mol/L-0.7mol=0.1mol,故n(Al
3+)=0.1mol,根據(jù)電荷守恒可知,n(Mg
2+)=
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=
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=0.2mol,故x=
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=0.1,故答案為:0.1.
分析:(1)令溶液中Al
3+的物質(zhì)的量為a mol,利用電荷守恒用a表示出Mg
2+的物質(zhì)的量,溶液中離子總的物質(zhì)的量為三種離子之和,也等于
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mol,據(jù)此列等式計算a的值(用x表示),利用極限法可知0<a<
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,據(jù)此計算x的取值范圍;
(2)溶液中的陽離子恰好完全沉淀,此時溶液中溶質(zhì)為KCl,根據(jù)守恒可知n(KCl)=n(KOH)=n(Cl
-),據(jù)此計算KOH溶液是體積;
(3)當(dāng)100 mL KOH剛好完全消耗,且Al
3+全部轉(zhuǎn)化為AlO
2-時,此時溶液中溶質(zhì)為KCl、KAlO
2,根據(jù)守恒可知n(K
+)=n(Cl
-)+n(AlO
2-),n(AlO
2-)=n(Al
3+),利用電荷守恒計算n(Mg
2+),據(jù)此計算x的值.
點(diǎn)評:本題考查混合物的有關(guān)計算,題目難度中等,注意利用守恒進(jìn)行的計算,(1)為易錯點(diǎn),可以直接利用極限法分析判斷x的最大值不能取等號.