2026年鲁教版适配小升初化学三模押题卷化学方程式与计算标准试卷第249套（含答案解析与可打印作答区）

小升初化学三模押题卷·方程式与计算·第249套请在指定区域内作答，保持卷面整洁。2026年鲁教版适配小升初化学三模押题卷化学方程式与计算标准试卷第249套（含答案解析与可打印作答区）学校：________________班级：________________姓名：________________考号：________________考试时间：120分钟满分：120分注意事项1.本卷共28题，满分120分。请先通读全卷，再按题号顺序作答。2.选择题每题只有一个最佳答案，请将答案填入答题栏；非选择题在题后指定空白区作答。3.计算题需写出化学方程式、已知量与未知量、比例关系、代入过程、单位和结论。4.相对原子质量：H-1，C-12，O-16，Na-23，Mg-24，Cl-35.5，K-39，Ca-40，Mn-55，Fe-56，Cu-64，Zn-65，S-32。5.诚信应考，独立完成；书写要清楚，数据保留按题目要求执行，未说明时计算结果保留一位小数或三位有效数字。一、选择题（本大题共15小题，每小题2分，共30分。每小题只有一个最佳答案）选择题答题栏题号123456789101112131415答案1.（2分）下列化学方程式书写正确，且符合质量守恒定律的是A.Mg+O₂→MgOB.2Mg+O₂→2MgOC.Mg₂+O₂→2MgOD.2Mg+O₂→Mg₂O₂2.（2分）实验室用过氧化氢溶液制取氧气时，下列说法正确的是A.二氧化锰作反应物，反应后质量必然减少B.该反应的化学方程式为2H₂O₂→2H₂O+O₂↑C.收集到的氧气能使带火星木条复燃D.反应前后原子种类改变3.（2分）在化学方程式2H₂+O₂→2H₂O中，参加反应的氢气和氧气的质量比为A.1:8B.1:4C.2:1D.4:14.（2分）下列有关化学方程式“Fe+CuSO₄→FeSO₄+Cu”的叙述，不正确的是A.铁能置换出铜B.反应前后铜元素守恒C.反应前后溶液颜色可能发生改变D.1个铁原子生成2个铜原子5.（2分）红磷在氧气中燃烧：4P+5O₂→2P₂O₅。若有12.4g红磷完全反应，理论上消耗氧气的质量为A.8.0gB.12.8gC.16.0gD.20.0g6.（2分）碳酸钙与足量稀盐酸反应：CaCO₃+2HCl→CaCl₂+H₂O+CO₂↑。若生成4.4g二氧化碳，则参加反应的碳酸钙质量为A.5.0gB.10.0gC.22.0gD.44.0g7.（2分）将少量白磷密封在充满氧气的玻璃管中点燃，冷却后称量，反应前后玻璃管及内容物总质量A.变大，因为生成固体B.变小，因为消耗氧气C.不变，因为装置密闭D.无法判断，因为白磷有毒8.（2分）配平反应Fe₂O₃+CO→Fe+CO₂后，CO₂前的化学计量数为A.1B.2C.3D.49.（2分）某石灰石样品12.0g与足量稀盐酸充分反应，生成CO₂4.4g。样品中CaCO₃的质量分数约为A.41.7%B.50.0%C.83.3%D.91.7%10.（2分）同温同压下，氢气和氯气反应生成氯化氢：H₂+Cl₂→2HCl。恰好完全反应时，氢气与氯气体积比为A.1:1B.1:2C.2:1D.2:311.（2分）下列现象最能说明有气体生成的是A.溶液由蓝色变为浅绿色B.试管壁变热C.有大量气泡逸出并能使澄清石灰水变浑浊D.固体逐渐溶解12.（2分）下列反应中，属于分解反应的是A.Zn+H₂SO₄→ZnSO₄+H₂↑B.CaO+H₂O→Ca(OH)₂C.2KMnO₄→K₂MnO₄+MnO₂+O₂↑D.NaOH+HCl→NaCl+H₂O13.（2分）将80g质量分数为10%的氯化钠溶液加水稀释到200g，稀释后溶液中氯化钠的质量分数为A.2%B.4%C.5%D.8%14.（2分）碳酸钠与稀盐酸反应：Na₂CO₃+2HCl→2NaCl+H₂O+CO₂↑。10.6g碳酸钠完全反应，理论上生成CO₂的质量为A.2.2gB.4.4gC.8.8gD.10.6g15.（2分）做“碳酸钙与稀盐酸反应前后质量变化”实验时，若在敞口烧杯中进行并直接称量，反应后总质量常变小，主要原因是A.反应不遵守质量守恒定律B.生成的二氧化碳逸散到空气中C.盐酸被完全消耗D.烧杯吸收空气中水蒸气二、填空与基础计算题（本大题共5小题，每小题6分，共30分）16.（6分）按要求完成下列化学方程式，并在括号内写出反应基本类型：

（1）镁条与稀盐酸反应：____Mg+____HCl→____MgCl₂+____H₂↑（________反应）

（2）氯酸钾受热分解：____KClO₃→____KCl+____O₂↑（________反应）

（3）氢氧化钠溶液与稀盐酸反应：NaOH+HCl→__________+__________。作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________17.（6分）镁在氧气中燃烧生成氧化镁。现有2.4g镁完全反应，生成氧化镁4.0g。请回答：参加反应的氧气质量为________g；镁与氧气的质量比为________；若生成8.0g氧化镁，理论上需要镁________g。作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________18.（6分）把一根洁净铁钉放入硫酸铜溶液中，一段时间后取出。该反应的化学方程式为____________________________；观察到铁钉表面有________色固体生成，溶液颜色由蓝色逐渐变为________色。作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________19.（6分）将80g质量分数为10%的氯化钠溶液加水稀释至200g。原溶液中氯化钠质量为________g；加入水的质量为________g；稀释后溶液的质量分数为________。作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________20.（6分）锌与稀盐酸反应：Zn+2HCl→ZnCl₂+H₂↑。13.0g锌完全反应，理论上生成氢气________g；参加反应的HCl质量为________g；若恰好用去100.0g稀盐酸，则该稀盐酸中HCl的质量分数为________。作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________三、材料、实验与图表分析题（本大题共4小题，每小题8分，共32分）21.（8分）某小组用高锰酸钾制取氧气。已知反应为2KMnO₄→K₂MnO₄+MnO₂+O₂↑。他们取31.6g高锰酸钾充分加热，记录如下信息。项目反应前固体质量反应后固体剩余质量收集气体情况数据31.6g28.4g排水法收集，集气瓶内气体能使带火星木条复燃（1）该反应属于________反应。

（2）根据质量守恒定律，生成氧气质量为________g。

（3）31.6gKMnO₄的物质转化中，理论上生成氧气多少克？请用化学方程式列式计算。

（4）若试管口没有稍向下倾斜，可能造成的后果是什么？作答区：____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________22.（8分）为测定某石灰石样品中碳酸钙的质量分数，取12.5g样品放入烧杯，加入足量稀盐酸，反应过程中烧杯及其中物质总质量变化如下表。杂质不与稀盐酸反应，不溶于水。反应时间/min02468总质量/g112.5110.9109.2108.1108.1（1）反应完全时生成CO₂的质量为________g。

（2）写出反应的化学方程式。

（3）计算样品中CaCO₃的质量分数。

（4）说明6min后总质量不再变化的原因。作答区：____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________23.（8分）阅读材料：水垢的主要成分之一是碳酸钙。家庭除垢剂中常含有酸性物质，能与碳酸钙反应生成可溶性盐、水和二氧化碳。实验小组用质量分数为10.0%的稀盐酸处理含CaCO₃90.0%的水垢样品5.0g，假设其他成分不与盐酸反应。（1）写出CaCO₃与稀盐酸反应的化学方程式。

（2）反应时出现气泡，检验该气体可选用的试剂是________。

（3）计算完全除去样品中CaCO₃至少需要10.0%的稀盐酸多少克。

（4）实际操作中常需稍过量加入除垢剂，原因是什么？作答区：____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________24.（8分）某同学测定纯碱样品中Na₂CO₃的质量分数。取样品12.0g，加水完全溶解，向溶液中加入足量CaCl₂溶液，过滤、洗涤、干燥后得到CaCO₃沉淀10.0g。反应为Na₂CO₃+CaCl₂→CaCO₃↓+2NaCl。（1）该实验中CaCl₂溶液需“足量”的目的是________________。

（2）生成CaCO₃沉淀10.0g，对应的Na₂CO₃质量是多少？

（3）计算样品中Na₂CO₃的质量分数。

（4）若沉淀未充分干燥就称量，会使计算出的Na₂CO₃质量分数偏大还是偏小？说明理由。作答区：____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________四、综合计算题（本大题共4小题，每小题7分，共28分）25.（7分）某补铁剂中含有铁粉和不与稀硫酸反应的辅料。取11.2g样品，加入足量稀硫酸，完全反应后收集到氢气0.36g。反应为Fe+H₂SO₄→FeSO₄+H₂↑。

（1）计算样品中铁的质量。

（2）计算该补铁剂中铁的质量分数。

（3）计算理论上生成FeSO₄的质量。作答区：____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________26.（7分）用氢气还原氧化铜：CuO+H₂→Cu+H₂O。某实验中取16.0gCuO充分反应。

（1）理论上可得到铜多少克？

（2）理论上生成水多少克？

（3）若实际得到铜12.0g，计算铜的产率。作答区：____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________27.（7分）某兴趣小组用68.0g质量分数为5.0%的过氧化氢溶液制取氧气，二氧化锰作催化剂。反应为2H₂O₂→2H₂O+O₂↑，假设H₂O₂完全分解且氧气全部逸出。

（1）溶液中H₂O₂的质量是多少？

（2）理论上生成氧气多少克？

（3）反应后液体质量是多少克？作答区：____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________28.（7分）50.0g氢氧化钠溶液恰好与36.5g质量分数为10.0%的稀盐酸完全反应。反应为NaOH+HCl→NaCl+H₂O。

（1）计算氢氧化钠溶液中NaOH的质量。

（2）计算该氢氧化钠溶液的质量分数。

（3）计算反应后所得溶液中NaCl的质量分数。作答区：____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

参考答案与解析评分说明：计算题过程分包含化学方程式、比例关系、代入、单位与结论；若结果正确但过程缺失，应酌情扣除过程分。1.B。解析：镁与氧气反应生成氧化镁，正确配平后为2Mg+O₂→2MgO，反应前后镁原子、氧原子个数均相等。A未配平，C、D中化学式不符合常见化合价。2.C。解析：过氧化氢分解生成水和氧气，二氧化锰是催化剂，反应前后质量和化学性质不变。化学方程式需写催化剂条件，收集到的氧气可使带火星木条复燃，反应前后原子种类不变。3.A。解析：2H₂与O₂反应时，氢气质量为4份，氧气质量为32份，质量比为4:32=1:8。化学计量数比不等于质量比，需乘相对分子质量。4.D。解析：Fe+CuSO₄→FeSO₄+Cu表示1个铁原子置换出1个铜原子，元素种类和原子个数守恒。硫酸铜溶液由蓝色逐渐变浅，生成铜单质。5.C。解析：4P+5O₂→2P₂O₅中，124gP消耗160gO₂。12.4gP消耗氧气质量为12.4×160÷124=16.0g。6.B。解析：CaCO₃与CO₂的质量关系为100:44。生成4.4gCO₂对应CaCO₃质量为4.4×100÷44=10.0g。7.C。解析：密闭装置中反应前后物质没有逸散，虽然白磷和氧气转化为五氧化二磷，但总质量不变，体现质量守恒定律。8.C。解析：配平为Fe₂O₃+3CO→2Fe+3CO₂，CO₂前的化学计量数为3。配平时先配铁，再配碳和氧。9.C。解析：4.4gCO₂对应CaCO₃10.0g，样品质量12.0g，质量分数为10.0÷12.0×100%≈83.3%。10.A。解析：同温同压下气体体积比等于化学计量数比。H₂+Cl₂→2HCl中氢气与氯气计量数比为1:1，因此体积比为1:1。11.C。解析：大量气泡说明有气体逸出，能使澄清石灰水变浑浊说明该气体含CO₂。颜色变化、放热、固体溶解本身不能单独证明气体生成。12.C。解析：分解反应是“一种物质生成两种或两种以上物质”的反应。2KMnO₄受热生成K₂MnO₄、MnO₂和O