高考物質(zhì)轉(zhuǎn)化真題及答案

一、單項(xiàng)選擇題1.下列物質(zhì)間的轉(zhuǎn)化在給定條件下能實(shí)現(xiàn)的是（）A.\(SiO_{2}\xrightarrow{HCl(aq)}SiCl_{4}\xrightarrow{H_{2}/高溫}Si\)B.\(FeS_{2}\xrightarrow{O_{2}/高溫}SO_{2}\xrightarrow{H_{2}O_{2}}H_{2}SO_{4}\)C.\(N_{2}\xrightarrow{H_{2}/高溫、高壓、催化劑}NH_{3}\xrightarrow{O_{2}/催化劑、加熱}NO_{2}\)D.\(MgCO_{3}\xrightarrow{HCl(aq)}MgCl_{2}(aq)\xrightarrow{電解}Mg\)答案：B2.已知\(X\)、\(Y\)、\(Z\)、\(W\)均為中學(xué)化學(xué)中常見的單質(zhì)或化合物，它們之間的轉(zhuǎn)化關(guān)系如圖所示（部分產(chǎn)物已略去）。則\(W\)、\(X\)不可能是（）A.\(W\)為\(NaOH\)溶液，\(X\)為\(AlCl_{3}\)溶液B.\(W\)為\(CO_{2}\)，\(X\)為\(NaOH\)溶液C.\(W\)為\(Fe\)，\(X\)為\(HNO_{3}\)D.\(W\)為\(O_{2}\)，\(X\)為\(C\)答案：C3.下列物質(zhì)的轉(zhuǎn)化在一定條件下不能一步實(shí)現(xiàn)的是（）A.\(Ca(ClO)_{2}\rightarrowHClO\)B.\(Fe_{2}O_{3}\rightarrowFe(OH)_{3}\)C.\(Na_{2}CO_{3}\rightarrowNaHCO_{3}\)D.\(CuO\rightarrowCu(NO_{3})_{2}\)答案：B4.下列物質(zhì)間的每一步轉(zhuǎn)化不能通過一步反應(yīng)實(shí)現(xiàn)的是（）A.\(Na_{2}CO_{3}\rightarrowCO_{2}\rightarrowNaHCO_{3}\)B.\(Fe\rightarrowFeCl_{2}\rightarrowFe(OH)_{2}\rightarrowFe(OH)_{3}\)C.\(Si\rightarrowSiO_{2}\rightarrowH_{2}SiO_{3}\)D.\(Na\rightarrowNa_{2}O_{2}\rightarrowNa_{2}CO_{3}\)答案：C5.下列物質(zhì)轉(zhuǎn)化在給定條件下能實(shí)現(xiàn)的是（）①\(Al_{2}O_{3}\xrightarrow{NaOH(aq)}NaAlO_{2}(aq)\xrightarrow{CO_{2}}Al(OH)_{3}\)②\(S\xrightarrow{O_{2}/點(diǎn)燃}SO_{3}\xrightarrow{H_{2}O}H_{2}SO_{4}\)③\(飽和NaCl溶液\xrightarrow{NH_{3}、CO_{2}}NaHCO_{3}\xrightarrow{\triangle}Na_{2}CO_{3}\)④\(Fe_{2}O_{3}\xrightarrow{HCl(aq)}FeCl_{3}(aq)\xrightarrow{\triangle}無水FeCl_{3}\)A.①③B.②③C.②④D.①④答案：A6.甲、乙、丙、丁四種物質(zhì)中，甲、乙、丙均含有相同的某種元素，它們之間具有如下轉(zhuǎn)化關(guān)系：甲\(\xrightarrow{丁}\)乙\(\underset{甲}{\stackrel{丁}{\rightleftharpoons}}\)丙。下列有關(guān)物質(zhì)的推斷不正確的是（）A.若甲為焦炭，則丁可能是\(O_{2}\)B.若甲為\(SO_{2}\)，則丁可能是氨水C.若甲為\(Fe\)，則丁可能是鹽酸D.若甲為\(NaOH\)溶液，則丁可能是\(CO_{2}\)答案：C7.能實(shí)現(xiàn)下列物質(zhì)間直接轉(zhuǎn)化的元素是（）單質(zhì)\(\xrightarrow{+O_{2}}\)氧化物\(\xrightarrow{+H_{2}O}\)酸或堿\(\xrightarrow{+NaOH或HCl}\)鹽A.硅B.硫C.銅D.鐵答案：B8.下列物質(zhì)的轉(zhuǎn)化在給定條件下能實(shí)現(xiàn)的是（）①\(Fe\xrightarrow{O_{2}/點(diǎn)燃}Fe_{2}O_{3}\xrightarrow{H_{2}SO_{4}}Fe_{2}(SO_{4})_{3}\)②\(SiO_{2}\xrightarrow{NaOH}Na_{2}SiO_{3}\xrightarrow{HCl}H_{2}SiO_{3}\)③\(AlCl_{3}\xrightarrow{電解}Al\xrightarrow{NaOH}NaAlO_{2}\)④\(Na\xrightarrow{O_{2}/點(diǎn)燃}Na_{2}O_{2}\xrightarrow{CO_{2}}Na_{2}CO_{3}\)A.①②B.②③C.②④D.③④答案：C9.已知\(A\)、\(B\)為單質(zhì)，\(C\)為化合物。能實(shí)現(xiàn)下述轉(zhuǎn)化關(guān)系的是（）\(A+B\xrightarrow{點(diǎn)燃}C\xrightarrow{溶于水}\)所得溶液\(\xrightarrow{惰性電極電解}A+B\)①若\(C\)溶于水后得到強(qiáng)堿溶液，則\(A\)可能是\(Na\)②若\(C\)溶液遇\(Na_{2}CO_{3}\)放出\(CO_{2}\)氣體，則\(A\)可能是\(H_{2}\)③若\(C\)溶液中滴加\(KSCN\)溶液顯紅色，則\(B\)可能為\(Fe\)④若\(C\)溶液中滴加\(NaOH\)溶液有藍(lán)色沉淀生成，則\(B\)可能為\(Cu\)A.①②B.②④C.①③D.③④答案：B10.下列物質(zhì)的轉(zhuǎn)化在給定條件下能實(shí)現(xiàn)的是（）A.\(NH_{3}\xrightarrow{O_{2}/催化劑、加熱}NO_{2}\xrightarrow{H_{2}O}HNO_{3}\)B.\(AlCl_{3}(aq)\xrightarrow{氨水}Al(OH)_{3}\xrightarrow{電解}Al\)C.\(Fe_{2}O_{3}\xrightarrow{HCl(aq)}FeCl_{3}(aq)\xrightarrow{\triangle}FeCl_{3}(s)\)D.\(MgCl_{2}(aq)\xrightarrow{石灰乳}Mg(OH)_{2}\xrightarrow{煅燒}MgO\)答案：D二、多項(xiàng)選擇題1.下列物質(zhì)轉(zhuǎn)化過程能一步實(shí)現(xiàn)的有（）A.\(Na_{2}O_{2}\rightarrowNaOH\)B.\(SiO_{2}\rightarrowH_{2}SiO_{3}\)C.\(Fe(OH)_{2}\rightarrowFe(OH)_{3}\)D.\(CuO\rightarrowCu(OH)_{2}\)答案：AC2.下列物質(zhì)間的轉(zhuǎn)化可以通過多個反應(yīng)一步實(shí)現(xiàn)的是（）A.\(Fe\rightarrowFeCl_{3}\)B.\(Si\rightarrowH_{2}SiO_{3}\)C.\(S\rightarrowH_{2}SO_{4}\)D.\(Al_{2}O_{3}\rightarrowAl(OH)_{3}\)答案：AC3.下列物質(zhì)轉(zhuǎn)化在一定條件下能夠?qū)崿F(xiàn)的是（）A.\(Cu\xrightarrow{濃HNO_{3}}Cu(NO_{3})_{2}\xrightarrow{NaOH溶液}Cu(OH)_{2}\)B.\(Al_{2}O_{3}\xrightarrow{NaOH溶液}NaAlO_{2}\xrightarrow{CO_{2}}Al(OH)_{3}\)C.\(Na_{2}O_{2}\xrightarrow{CO_{2}}Na_{2}CO_{3}\xrightarrow{Ca(OH)_{2}}NaOH\)D.\(Fe_{2}O_{3}\xrightarrow{HCl溶液}FeCl_{3}\xrightarrow{Cu}Fe\)答案：ABC4.以下物質(zhì)間的轉(zhuǎn)化能通過一步反應(yīng)實(shí)現(xiàn)的有（）A.\(Al_{2}O_{3}\rightarrowAl(OH)_{3}\)B.\(Cu\rightarrowCu(NO_{3})_{2}\)C.\(Fe(OH)_{3}\rightarrowFe_{2}O_{3}\)D.\(Na_{2}CO_{3}\rightarrowNaHCO_{3}\)答案：BCD5.下列物質(zhì)轉(zhuǎn)化關(guān)系正確的是（）A.\(Al\xrightarrow{NaOH溶液}NaAlO_{2}\xrightarrow{過量CO_{2}}Al(OH)_{3}\)B.\(S\xrightarrow{O_{2}/點(diǎn)燃}SO_{3}\xrightarrow{H_{2}O}H_{2}SO_{4}\)C.\(N_{2}\xrightarrow{H_{2}/高溫、高壓、催化劑}NH_{3}\xrightarrow{O_{2}/催化劑、加熱}NO\)D.\(Fe_{2}O_{3}\xrightarrow{HCl溶液}FeCl_{3}\xrightarrow{Fe}FeCl_{2}\)答案：ACD6.下列物質(zhì)間的轉(zhuǎn)化在一定條件下可以實(shí)現(xiàn)的是（）A.\(C\xrightarrow{SiO_{2}/高溫}Si\xrightarrow{Cl_{2}/高溫}SiCl_{4}\xrightarrow{H_{2}/高溫}Si\)B.\(Na_{2}CO_{3}\xrightarrow{HCl}NaCl\xrightarrow{NH_{3}、CO_{2}、H_{2}O}NaHCO_{3}\)C.\(Fe_{2}O_{3}\xrightarrow{HCl}FeCl_{3}\xrightarrow{Cu}FeCl_{2}\)D.\(H_{2}S\xrightarrow{O_{2}/點(diǎn)燃}SO_{2}\xrightarrow{O_{2}/催化劑、加熱}SO_{3}\)答案：ABCD7.能實(shí)現(xiàn)下列物質(zhì)間直接轉(zhuǎn)化的是（）A.\(Al\rightarrowAl_{2}O_{3}\rightarrowAl(OH)_{3}\)B.\(Na\rightarrowNa_{2}O_{2}\rightarrowNa_{2}CO_{3}\)C.\(S\rightarrowSO_{2}\rightarrowH_{2}SO_{4}\)D.\(NH_{3}\rightarrowNO\rightarrowHNO_{3}\)答案：BCD8.下列物質(zhì)轉(zhuǎn)化過程合理的是（）A.\(MnO_{2}\xrightarrow{濃鹽酸/加熱}Cl_{2}\xrightarrow{石灰乳}Ca(ClO)_{2}\)B.\(Fe\xrightarrow{稀HNO_{3}}Fe(NO_{3})_{2}\xrightarrow{NaOH溶液}Fe(OH)_{2}\)C.\(MgCl_{2}(aq)\xrightarrow{加熱蒸干}MgCl_{2}(s)\xrightarrow{電解}Mg\)D.\(CuO\xrightarrow{H_{2}SO_{4}}CuSO_{4}(aq)\xrightarrow{NaOH溶液、葡萄糖溶液/加熱}Cu_{2}O\)答案：AD9.下列物質(zhì)間的轉(zhuǎn)化能一步實(shí)現(xiàn)的是（）A.\(Fe_{3}O_{4}\xrightarrow{Al/高溫}Fe\)B.\(Na_{2}O_{2}\xrightarrow{CO_{2}}Na_{2}CO_{3}\)C.\(SiO_{2}\xrightarrow{H_{2}O}H_{2}SiO_{3}\)D.\(NH_{4}Cl\xrightarrow{NaOH溶液/加熱}NH_{3}\)答案：ABD10.下列物質(zhì)轉(zhuǎn)化在給定條件下可行的是（）A.\(Fe_{2}O_{3}\xrightarrow{HCl溶液}FeCl_{3}(aq)\xrightarrow{加熱蒸干}Fe_{2}O_{3}\)B.\(NaCl(aq)\xrightarrow{電解}Na\xrightarrow{O_{2}/點(diǎn)燃}Na_{2}O_{2}\)C.\(Mg(OH)_{2}\xrightarrow{HCl溶液}MgCl_{2}(aq)\xrightarrow{電解}Mg\)D.\(Al_{2}O_{3}\xrightarrow{NaOH溶液}NaAlO_{2}(aq)\xrightarrow{CO_{2}}Al(OH)_{3}\)答案：AD三、判斷題1.\(Na_{2}O\)與\(Na_{2}O_{2}\)之間可以相互轉(zhuǎn)化。（√）2.\(SiO_{2}\)能一步轉(zhuǎn)化為\(H_{2}SiO_{3}\)。（×）3.\(Fe(OH)_{2}\)在空氣中加熱可以直接得到\(Fe_{2}O_{3}\)。（√）4.\(CuO\)可以通過與水反應(yīng)生成\(Cu(OH)_{2}\)。（×）5.\(NH_{3}\)催化氧化可以直接生成\(NO_{2}\)。（×）6.\(Al_{2}O_{3}\)既能與酸反應(yīng)又能與堿反應(yīng)，所以它是兩性氧化物，\(Al(OH)_{3}\)也具有兩性。（√）7.\(S\)在足量\(O_{2}\)中燃燒可以直接生成\(SO_{3}\)。（×）8.\(NaHCO_{3}\)受熱分解可以轉(zhuǎn)化為\(Na_{2}CO_{3}\)。（√）9.\(FeCl_{3}\)溶液與\(Fe\)反應(yīng)可以實(shí)現(xiàn)\(FeCl_{3}\)到\(FeCl_{2}\)的轉(zhuǎn)化。（√）10.\(CaCO_{3}\)高溫分解可以得到\(CaO\)。（√）四、簡答題1.簡述\(Fe\)、\(FeCl_{2}\)、\(FeCl_{3}\)之間的相互轉(zhuǎn)化關(guān)系及相關(guān)反應(yīng)方程式。答案：\(Fe\)與\(Cl_{2}\)反應(yīng)生成\(FeCl_{3}\)：\(2Fe+3Cl_{2}\stackrel{點(diǎn)燃}{=\!=\!=}2FeCl_{3}\)；\(FeCl_{3}\)與\(Fe\)反應(yīng)轉(zhuǎn)化為\(FeCl_{2}\)：\(2FeCl_{3}+Fe=3FeCl_{2}\)；\(FeCl_{2}\)與\(Cl_{2}\)反應(yīng)又生成\(FeCl_{3}\)：\(2FeCl_{2}+Cl_{2}=2FeCl_{3}\)；\(Fe\)與鹽酸反應(yīng)生成\(FeCl_{2}\)：\(Fe+2HCl=FeCl_{2}+H_{2}\uparrow\)。2.寫出\(Al\)及其化合物之間轉(zhuǎn)化的常見反應(yīng)方程式（涉及\(Al\)、\(Al_{2}O_{3}\)、\(Al(OH)_{3}\)、\(NaAlO_{2}\)）。答案：\(Al\)與\(O_{2}\)反應(yīng)生成\(Al_{2}O_{3}\)：\(4Al+3O_{2}=2Al_{2}O_{3}\)；\(Al\)與\(NaOH\)溶液反應(yīng)生成\(NaAlO_