2026年鲁教版适配七年级化学期末模拟卷化学方程式与计算标准试卷第229套（含答案解析与可打印作答区）

2026年鲁教版适配七年级化学期末模拟卷化学方程式与计算标准试卷第229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分答题说明：1.本试卷共28题，所有题目均围绕化学方程式、质量守恒、物质组成与计算展开。2.选择题只有一个最佳答案，请将答案填入答题栏；非选择题请在题后作答区书写过程。3.计算题要写出化学方程式、相对分子质量关系、代入过程、单位和结论。4.保持卷面整洁，答案写在规定区域内。诚信提示：独立完成，认真审题，规范书写化学式和化学方程式。题型题号分值考查重点一、单项选择题1—1236分化学变化、方程式意义、配平与基础计算二、填空与基础应用题13—2032分化学式书写、质量关系、守恒分析三、材料与实验探究题21—2428分实验现象、数据处理、反应表达四、综合计算题25—2824分方程式计算、样品纯度、综合推断一、单项选择题（本题共12小题，每小题3分，共36分。每小题只有一个最佳答案）选择题答题栏：1234567891011121.（3分）下列变化中，能够用化学方程式表示且属于化学变化的是（）。A.冰块融化成水B.铁丝在氧气中燃烧生成黑色固体C.酒精挥发D.玻璃被切割成小块2.（3分）关于化学方程式2H₂+O₂→2H₂O的说法，正确的是（）。A.氢气和氧气在任意条件下都能生成水B.每2个氢分子和1个氧分子反应生成2个水分子C.反应前后分子种类不变D.反应前后氧元素的质量减少3.（3分）配平化学方程式Fe+O₂→Fe₃O₄后，铁和氧气前的化学计量数依次为（）。A.1，1B.2，1C.3，2D.3，44.（3分）在密闭容器中加热一定质量的铜粉，铜与氧气反应生成氧化铜。冷却后称量容器及其中物质的总质量，结果应为（）。A.一定增大B.一定减小C.保持不变D.先增大后减小5.（3分）下列化学方程式书写正确的是（）。A.Mg+O₂→MgO₂B.C+O₂→CO₂C.H₂O→H₂+OD.CaCO₃→CaO+CO₂（没有条件也一定发生）6.（3分）已知相对原子质量：H-1，O-16。水中氢元素与氧元素的质量比为（）。A.1:8B.1:16C.2:1D.8:17.（3分）某反应的文字表达式为“碳酸钙+盐酸→氯化钙+水+二氧化碳”。下列判断正确的是（）。A.反应物有3种B.生成物有3种C.二氧化碳是反应物D.该反应不符合质量守恒定律8.（3分）镁在氧气中燃烧的化学方程式为2Mg+O₂→2MgO。若24g镁完全反应，需要氧气的质量为（）。A.8gB.16gC.24gD.40g9.（3分）判断一个化学方程式是否配平，最直接的依据是（）。A.反应条件是否写出B.反应前后各元素原子个数是否相等C.生成物是否有气体符号D.化学式是否都很复杂10.（3分）在实验室用石灰石和稀盐酸制取二氧化碳时，下列做法合理的是（）。A.用燃着的木条检验二氧化碳B.用澄清石灰水检验二氧化碳C.把反应装置敞口放在天平上验证质量守恒D.用向上排空气法收集所有气体11.（3分）将10gA与足量B反应，生成18gC和4gD，则参加反应的B的质量为（）。A.8gB.12gC.14gD.32g12.（3分）下列关于化学方程式计算的说法中，错误的是（）。A.计算前应先检查方程式是否配平B.质量关系来自化学方程式中各物质的相对分子质量和计量数C.题中给出的任意质量都可以直接代入，不必判断是否完全反应D.结果要带单位，并结合题意作答二、填空与基础应用题（本题共8小题，每小题4分，共32分）13.（4分）按要求写出下列反应的化学方程式，并注明必要的反应条件。（1）碳在充足氧气中燃烧：__________________________________________（2）氢气在氧气中燃烧：__________________________________________（3）镁条在氧气中燃烧：__________________________________________（4）加热高锰酸钾制取氧气：______________________________________14.（4分）在横线上填入适当的化学计量数，使下列化学方程式配平。（1）____Al+____O₂→____Al₂O₃（2）____P+____O₂→____P₂O₅（3）____H₂O₂→____H₂O+____O₂↑（4）____CO+____O₂→____CO₂15.（4分）某同学在密闭装置中让4.8g镁粉充分燃烧，生成的氧化镁质量为8.0g。请依据质量守恒定律填空。（1）参加反应的氧气质量为__________g。（2）该反应中镁、氧气、氧化镁的质量比为________________。（3）化学方程式为__________________________________________。（4）若改在敞口坩埚中称量，生成固体质量可能偏小，原因是__________________________________________。16.（4分）根据相对原子质量H-1、C-12、O-16、Ca-40，完成计算。（1）CaCO₃的相对分子质量为__________。（2）CO₂的相对分子质量为__________。（3）CaCO₃中钙元素、碳元素、氧元素的质量比为________________。（4）50gCaCO₃中含氧元素的质量为__________g。17.（4分）实验室常用大理石或石灰石与稀盐酸反应制取二氧化碳。（1）写出反应的化学方程式：________________________________________________________。（2）反应中产生的气体能使澄清石灰水________________。（3）若收集到二氧化碳的质量偏小，可能的实验原因是________________________________。（4）从元素守恒角度看，生成的二氧化碳中的碳元素来自________________。18.（4分）下表是某反应前后物质质量的记录，请填写空缺并判断反应类型。物质甲乙丙丁反应前质量/g20600反应后质量/g2018x（1）x=__________。（2）甲减少的质量为__________g，乙减少的质量为__________g。（3）丙和丁属于________________物。（4）该反应可概括为“甲+乙→丙+丁”，判断依据是________________________________。19.（4分）将一定质量的氯酸钾和少量二氧化锰混合加热，完全反应后得到氧气和氯化钾。已知二氧化锰作催化剂。（1）反应的化学方程式为________________________________________________________。（2）二氧化锰在反应前后质量__________，化学性质__________。（3）若反应前混合物为12.0g，反应后固体为8.16g，则生成氧气质量为__________g。（4）计算氧气质量时运用的基本规律是________________。20.（4分）把下列文字信息转化为规范表达，并完成对应问题。铁丝在氧气中剧烈燃烧，火星四射，生成黑色固体四氧化三铁。（1）化学方程式：________________________________________________________。（2）反应现象中的“黑色固体”是________________。（3）若有5.6g铁完全反应，理论上生成四氧化三铁质量为__________g。（4）实际实验中常在集气瓶底部放少量水或细沙，目的是________________________________。三、材料与实验探究题（本题共4小题，每小题7分，共28分）21.（7分）某小组用白磷燃烧实验探究质量守恒定律。装置为带玻璃管的锥形瓶，瓶内放少量白磷，玻璃管末端套一只气球。实验前称量整套装置质量为m₁；加热使白磷燃烧，待装置冷却后再次称量，质量为m₂。请回答：（1）白磷燃烧生成五氧化二磷，化学方程式为________________________________。（2分）（2）实验中观察到气球先膨胀后变瘪，原因是________________________________________。（2分）（3）理论上m₁与m₂的关系为________________，依据是________________。（2分）（4）若冷却前就称量，读数可能不稳定，请从气体体积角度说明原因：________________________。（1分）________________________________________________________________________________________________________________________________________________________________22.（7分）某实验小组制取并检验氧气，记录如下：取一定量高锰酸钾放入试管，加热；用排水法收集一瓶气体；把带火星的木条伸入集气瓶中，木条复燃。实验步骤现象或数据加热前固体质量15.8g充分加热后剩余固体质量14.2g气体检验带火星木条复燃（1）高锰酸钾分解的化学方程式为________________________________________。（2分）（2）收集到的气体是__________，检验依据是________________________________。（2分）（3）由质量差可知生成气体质量为__________g。（1分）（4）若试管口未略向下倾斜，可能造成的后果是________________________________。（1分）（5）反应后固体中除二氧化锰外，还含有________________。（1分）________________________________________________________________________________________________________________________________________________________________23.（7分）阅读材料并回答问题。厨房中常见的小苏打主要成分为碳酸氢钠，遇到酸性物质会产生二氧化碳。某兴趣小组将8.4g碳酸氢钠与足量稀盐酸反应，反应后测得生成二氧化碳4.4g。反应可表示为：NaHCO₃+HCl→NaCl+H₂O+CO₂↑。（1）该化学方程式是否已配平？__________，判断理由是________________________________。（2分）（2）从反应物角度看，二氧化碳中的碳元素来自________________。（1分）（3）若用密闭装置反应，反应前后总质量应________________。（1分）（4）根据相对原子质量Na-23、H-1、C-12、O-16，计算8.4gNaHCO₃完全反应理论生成CO₂的质量，并与材料数据比较。（3分）________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________24.（7分）某校劳动实践活动中，学生观察到铁制工具在潮湿空气中容易生锈。为探究铁生锈是否消耗氧气，小组将湿润铁粉放入密闭量筒装置中，记录水面上升情况。时间/min010203040量筒内气体体积/mL10092868280（1）量筒内气体体积减小，说明铁生锈过程中消耗了________________。（1分）（2）40min时气体减少了__________mL，占初始气体体积的__________%。（2分）（3）若空气中氧气约占体积的21%，该实验数据与理论值接近，说明装置气密性________________。（1分）（4）铁锈主要成分可看作含水氧化铁，铁生锈属于________________变化。（1分）（5）请提出一条防止铁制品生锈的措施，并说明理由：________________________________。（2分）________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________四、综合计算题（本题共4小题，每小题6分，共24分）25.（6分）将4.90g氯酸钾在二氧化锰催化下充分加热，发生反应：2KClO₃→2KCl+3O₂↑。已知相对原子质量K-39、Cl-35.5、O-16。计算：（1）理论上生成氧气的质量。（3分）（2）反应后生成氯化钾的质量。（2分）（3）二氧化锰的质量是否计入反应生成物质量？说明理由。（1分）________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________26.（6分）某石灰石样品10.0g与足量稀盐酸完全反应，生成二氧化碳3.52g。样品中的杂质不与稀盐酸反应。反应为：CaCO₃+2HCl→CaCl₂+H₂O+CO₂↑。已知相对原子质量Ca-40、C-12、O-16。计算：（1）参加反应的碳酸钙质量。（3分）（2）该石灰石样品中碳酸钙的质量分数。（2分）（3）若装置漏气，测得二氧化碳质量偏小，则计算出的质量分数将偏大还是偏小？（1分）________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________27.（6分）水电解实验中，电解水的化学方程式为2H₂O→2H₂↑+O₂↑。若完全分解18.0g水，按化学方程式计算：（1）生成氢气的质量。（2分）（2）生成氧气的质量。（2分）（3）从微观角度说明电解水与水蒸发的本质区别。（2分）________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________28.（6分）在硫酸铜溶液中加入5.6g铁粉，铁与硫酸铜恰好完全反应，反应为：Fe+CuSO₄→FeSO₄+Cu。已知相对原子质量Fe-56、Cu-64。计算：（1）理论上生成铜的质量。（3分）（2）反应后固体质量比加入的铁粉质量增加多少克？（1分）（3）从元素守恒角度说明生成铜的来源。（2分）________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________可打印作答续页（用于第21—28题补充过程）需要补充实验现象、计算步骤、单位换算或结论说明时，可在本页继续作答。______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________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参考答案与解析本部分按题号逐题给出答案、关键过程和失分提醒。1.答案：B。解析：铁丝在氧气中燃烧有新物质四氧化三铁生成，属于化学变化，可用化学方程式表示。冰融化、酒精挥发和玻璃切割只是状态或形状变化，没有新物质生成。2.答案：B。解析：方程式中计量数表示微粒个数关系：2个氢分子和1个氧分子反应生成2个水分子。反应需要点燃等条件；反应前后分子种类改变，但元素种类和元素质量守恒。3.答案：C。解析：配平后为3Fe+2O₂→Fe₃O₄。反应前铁原子3个、氧原子4个，反应后Fe₃O₄中铁原子3个、氧原子4个，原子个数相等。4.答案：C。解析：在密闭容器中，铜粉和氧气反应生成氧化铜，参与反应的氧气仍在容器内，总质量不因化学反应而改变，符合质量守恒定律。5.答案：B。解析：碳在氧气中充分燃烧生成二氧化碳，C+O₂→CO₂配平且化学式正确。镁燃烧应为2Mg+O₂→2MgO；水分解生成H₂和O₂；碳酸钙分解需要高温条件。6.答案：A。解析：H₂O中氢元素质量为2×1=2，氧元素质量为16，质量比为2:16=1:8。7.答案：B。解析：反应物是碳酸钙和盐酸，共2种；生成物是氯化钙、水、二氧化碳，共3种。所有化学反应都遵守质量守恒定律。8.答案：B。解析：2Mg+O₂→2MgO中48g镁对应32g氧气，24g镁对应16g氧气。计算时要用配平后的质量关系。9.答案：B。解析：化学方程式配平的本质是使反应前后各元素原子