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1、中考_化学:初中化学计算题题型汇总中考 化学:初中化学计算题题型汇总 1、生铁和钢都是铁合金,生铁中碳的含量在2.0%4.3%之间,钢中碳的含量在0.03%2%之间。将一块质量为10.0g的铁合金放入锥形瓶中,再向锥形瓶中加入100g稀H2SO4,恰好使铁合金中的铁完全反应(碳不熔于稀硫酸;铁合金中其他元素含量很低,可忽略不计),测得生成H2的体积为4.00L(H2在该条件下的密度为0.0880g/L)。试根据计算回答:(计算结果保留三位效数字) ()该铁合金是生铁还是钢, (2)反应后所得溶液的溶质质量分数。 2、将3.1克氧化钠加到质量为m的水中,完全反应后,得到溶质的质量分数为16%的溶
2、液,在该溶液中加入50克稀盐酸,两者恰好完全反应。(氧化钠与水反应的化学方程式为:Na2O+H2O=2NaOH)。求: (1) m的值 (2) 所加盐酸中溶质的质量分数。 (3) 若要使反应后的氧化钠溶液成为20C时的饱和溶液,至少要蒸发掉多少克水,(20C时,氧化钠的溶解度为36克) (1)m=21.9g (2)设盐酸中溶质的质量分数为y,生成NaCl的质量为a y=7.3% a=5.85g (3)设蒸发质量为b的水后可得到20?C时NaCl的饱和溶 5.85g/(50g+25g,5.85g,b)=36/100 b=52.9g 3、一个青少年正常情况下每天约需0.8g钙。若每天从食物中得到0
3、.6g钙,其余,由钙片补充,则每天需吃含葡萄糖酸钙(CHO)2Ca的质量分数为85%的钙片_g。(保留两位小数) 61174、为了对海水中的氯化物(假设以NaCl计算)进行成分分析,甲、乙、丙三位同学分别进行实验数据如下,请仔细观察分析,回答下列问题: 甲 乙 丙 所取海水样品的质量 /g 100 50 50 加入AgNO3溶液的质量 /g 50 50 75 反应后所得的沉淀物的质量 /g 1(435 1(435 1(435 (1) 两溶液恰好完全反应的是_的实验。 请计算海水中的氯化物(以氯化钠计算)的质量分数是多少, 5、“烟台苹果”享誉全国。波尔多液是烟台果农常用的一种果树杀菌农药。一果
4、农管理了5亩果园,准备为果树喷洒一次波尔多液,他现有8%的硫酸铜溶液30?、蓝矾(CuSO4?5H2O)10?,每亩地需要用150?1%的硫酸铜溶液来配制波尔多液。请你计算一下他现有的硫酸铜药品是否能够满足需要,(已知:蓝矾中CuSO4的质量分数为64%) 解:现有的药品中含硫酸铜的质量为: 30?8%+10?64%=8.8? 5亩果园需硫酸铜的质量为: 150?1%5=7.5? 因为8.8?7.5? 所以这一果农现有的硫酸铜药品能够满足需要 6、实验室现需1.12L纯净的氧气(标准状况下,氧气的密度是1.429g/L)。某同学用质量比为3:1的氯酸钾和二氧化锰制取氧气,并回收二氧化锰和氧化钾
5、。下表为该同学实验的有关数据: 实验收集1.12L氧气后试管内固体残余物的质量 充分加热后试管内固体残余物的质量 steel bolts, screws and studs GB/T3098.6-2000 scope of this project are: 1, 2, 3 aluminium sheet curtain wall of glass curtain wall and the Aluminum plastic Board curtain wall 4, and stone curtain wall 5, and curtain wall buried pieces 6, and
6、light steel points type glass rain canopy, and aluminum Board rain canopy 7, and aluminum Venetian 8, and outdoor advertising (containing roofing and the State surface advertising) 9 to spring glass door 10, and concrete structure and curtain wall between of layer between, and around fire seal block
7、ing 11, and anti-mine 12, and daughter wall top processing 13, and pan light ming, engineering of tie construction 14, and buildings of deformation sewing (settlement sewing, and expansion joints,) 15, and After the detailed design drawings issued a wall embedded part is inconsistent with the tender
8、er needs to increase fixed curtain wall structure 16, decorative surfaces and decorative surface and joints between parts of other buildings facing sealing process, required to support the outer decoration of a hole, hole, plugging holes and so on. This works more decorative items, in order to ensur
9、e that the decoration construction progress and quality, trying to avoid construction does not . Processing. After installation is complete if you have damaged galvanized parts should 7.98g 6.92g 问该同学最多能回收到多少克二氧化锰,(结果精确到0。01g) 在实际操作中,会损失少量氧气。 解:设原混合物中MnO质量为x,则KClO质量为3x 23? 2KClO=3KCl+3O32 245 149 3x
10、 6.92,x 245/3x=149/(6.92g,x) x=2.45g 答:最多可回收2.45g的二氧化锰。 7、将18g不纯的氯化铜样品(杂质不溶于水,也不参与反应)跟一定量的氢氧化钠恰好完全反应,得到溶质量分数为20%的溶液58.5g。求: (1) 样品中氯化铜的质量分数。 (2) 加入的氢氧化钠溶液的质量。 解:设样品中CuCl的质量为x,NaOH溶液中含NaOH质量为y 258.5g20%=11.7g CuCl+2NaOH=Cu(OH)?+2NaCl 22135 80 117 x y 1.17 (1)135:x=117:11.7g x=13.5g 样品中CuCl的质量分数为:(13.
11、5g/18g)100%=75% 2(2)80:y=117:11.7g y=80g NaOH溶液中的水的质量为58.5g,11.7g=46.8g 所以NaOH溶液的质量为:8g+46.8g=54.8g 8、一定质量10%的A溶液和一定质量10%的CaCl溶液恰好完全反应,生成白色沉淀B,化学方2程式为:A+CaCl=B?+2NaCl(已配平) 2(1)、A物质必定含有的金属元素是_,A和B的相对分子质量之差为_。 (2)、滤出沉淀B,加入足量稀盐酸,B逐渐溶解,收集到无色无气味的气体4.4g。求滤出B后所得溶液中NaCl的质量分数。 )钠(或Na) 6 (1(2)解:由题意判断:B是CaCO3,
12、A是Na2CO3 设生成CaCO的质量为w,NaCO溶液的质量为x,CaCl溶液的质量为y, 3232生成NaCl的质量为z CaCO+2HCl=CaCl+CO?+CO+HO 32222100 44 w 4.4g 100/44=w/4.4g w=10g NaCO+CaCl=CaCO?+2NaCl 2323106 111 100 117 10%x 10%y 10g z 106/100=10%x/10g x=106g ;111/100=10%y/10g y=111g ; 117/100=z/10g z=11.7g NaCl溶液中溶质的质量分数为:11.7g/(106g+111g,10g)100%=
13、5.7% 9、把15g不纯硫酸钠(所含杂质不溶于水),放入85.8g水中配成溶液。过滤后,取10g溶液加入适量的氯化钡溶液,恰好完全反应,生成2.33g白色沉淀。试计算:(最后结果保留二位小数) (,)10g溶液中硫酸钠的质量。 (,)原不纯硫酸钠中含硫酸钠的质量分数。 (1)解:设10g溶液中硫酸钠的质量为x NaSO+BaSO=BaSO+2NaCl 2444?ssing. After installation is complete if you have damaged galvanized parts shouldProce ve items, in order to ensure t
14、hat the decoration construction progress and quality, trying to avoid construction does not .sealing process, required to support the outer decoration of a hole, hole, plugging holes and so on. This works more decoratiacing ase fixed curtain wall structure 16, decorative surfaces and decorative surf
15、ace and joints between parts of other buildings fnts,) 15, and After the detailed design drawings issued a wall embedded part is inconsistent with the tenderer needs to increation sewing (settlement sewing, and expansion joimine 12, and daughter wall top processing 13, and pan light ming, engineerin
16、g of tie construction 14, and buildings of deform-urtain wall between of layer between, and around fire seal blocking 11, and antiand coutdoor advertising (containing roofing and the State surface advertising) 9 to spring glass door 10, and concrete structure uried pieces 6, and light steel points t
17、ype glass rain canopy, and aluminum Board rain canopy 7, and aluminum Venetian 8, andtain wall 4, and stone curtain wall 5, and curtain wall b2000 scope of this project are: 1, 2, 3 aluminium sheet curtain wall of glass curtain wall and the Aluminum plastic Board cur-steel bolts, screws and studs GB
18、/T3098.62 考试必备 142 233 x 2.33g 142/x=233/2.33g x=1.42g (2)85.8g水溶解的NaSO的质量为 241.42g/(10g,1.42g)85.8g=14.2g 原不纯硫酸钠中NaSO的质量为(14.2g/15g)100%=94.67% 2410、现有一种铜和氧化亚铁的混合物样品。为测定样品中氧化亚铁的含量,某学生取20.0g此样品置于烧瓶中,分四次加入密度为1.22g/cm?的硫酸并做相关记录。有关数据记录如下表: 加硫酸的序号 加入硫酸的体积/mL 剩余固体的质量/g 1 20(0 16(0 2 20(0 12(0 3 20(0 8(0
19、4 20(0 8(0 试计算:(1)10.0g此样品中铜的质量。 (2)样品中氧化亚铁的质量分数。 (,)所用硫酸溶液溶质的质量分数。 (1)4克。 (2)氧化亚铁的质量为10,4,6克 所以氧化亚铁的质量分数为6?10,0.6,60% (3)20毫升溶液溶解4克的氧化亚铁 FeO+H2SO4-FeSO4+H2O 72 98 4 x x=5.44 溶液的质量分数为5.44?(201.22),22.3% 11、青少年正处于生长发育时期,每天需要摄取足量的蛋白质,蛋白质的代谢产物主要是尿素CO(NH)。若从食物中摄取的蛋白质经体内新陈代谢后完全转化为尿素排出体外,每人每天相当22于排出尿素30g。
20、 (1)30g尿素中含氮元素多少克? (2)已知蛋白质中氮元素的平均质量分数为16,,则每人每天至少应从食物里摄取的蛋白质为多少克? 13、某炼铁厂用含氧化铁80%的赤铁矿冶炼生铁。若要炼出1120t含铁95%的生铁,需要含氧化铁80%的赤铁矿多少吨, Fe2O3-2Fe 160-112 80%X-1120t 解得X,2000吨 14、不纯的铁片5.6g与足量的HCl反应,放出0.21g氢气,则铁片中可能含有的一种金属杂质为 A(Zn B(Pb C(Cu D Mg 如果是纯的铁那放出的氢就是0.2, 但放出的大于0.2,那就含有相对原子量小于铁且要和盐酸反应的物质 15、将26g某金属混合物投
21、入到足量的稀硫酸中,共收集到2g氢气,该金属混合物的组成可能是zed parts shouldand quality, trying to avoid construction does not . Processing. After installation is complete if you have damaged galvaniress e, hole, plugging holes and so on. This works more decorative items, in order to ensure that the decoration construction prog
22、surface and joints between parts of other buildings facing sealing process, required to support the outer decoration of a hol ativeedded part is inconsistent with the tenderer needs to increase fixed curtain wall structure 16, decorative surfaces and decorof deformation sewing (settlement sewing, an
23、d expansion joints,) 15, and After the detailed design drawings issued a wall emb mine 12, and daughter wall top processing 13, and pan light ming, engineering of tie construction 14, and buildings-ng 11, and antiing) 9 to spring glass door 10, and concrete structure and curtain wall between of laye
24、r between, and around fire seal blockivertisd aluminum Board rain canopy 7, and aluminum Venetian 8, and outdoor advertising (containing roofing and the State surface adcurtain wall 4, and stone curtain wall 5, and curtain wall buried pieces 6, and light steel points type glass rain canopy, an2000 s
25、cope of this project are: 1, 2, 3 aluminium sheet curtain wall of glass curtain wall and the Aluminum plastic Board -steel bolts, screws and studs GB/T3098.63 ( ) A、Mg和Zn B、Fe和Zn C、Zn和Cu D、Cu和Fe 用计算的方法 24g Mg 生成2g 氢 64g Zn 生成2g 氢气 56g Fe 生成2g 氢气 铜不生成氢气 这里 26g 生成了2g 是金属混合的结果 所以必须一个大于26 一个小于26 小于26的在这
26、4中金属中 只有Mg 了 你会发现只有第一组有这个可能 16、A、B、C三种物质各15 g,它们充分反应只能生成30 g新物质D。若增加10 g C,A与C恰好完全反应,则参加反应的A与B的质量比为? A,B,C,D xy1530 1525 x/15,15/25 x,9 又因为x,y,15,30 所以:y,15,9,6 则:参加反应的A与B的质量比,x:y,9:6,3:2 18、在反应2A+5B=2C+4D中,C、D的相对分子质量之比为9?22(若2.6 gA与B完全反应后(生成8.8gD。则在此反应中B与D的质量比为( ) A(4?9 B(8?1 C 10 ? 11 D(31?44 19、阳
27、光牌小包装“脱氧剂”成分为Fe粉、活性炭及少量NaCl、水。使用一段时间后,其中的Fe粉会转变成FeO而变质。某化学兴趣小组欲探究使用过的阳光牌“脱氧剂”的变质程度(已23变质的Fe粉占变质前Fe粉的质量分数),设计并进行如下探究过程。 )取食品包装袋中的阳光牌“脱氧剂”一袋,将里面的固体溶于水,过滤、洗涤、步骤(1干燥滤渣。 步骤(2)取步骤(1)中的滤渣8.0 g,加入足量的稀HSO与滤渣充分反应,过滤、洗涤、24干燥得固体1.2 g。 步骤(3)取步骤(2)中的滤液,加入足量的NaOH溶液,得到的固体经洗涤后转移到坩埚中,充分加热、冷却、称量,得到8.0 g FeO(注:滤液中的Fe元素
28、已全部转化为FeO)。 2323求:(1)8.0 g滤渣中Fe和FeO两种物质的总质量。 23(2)该“脱氧剂”在未变质时,Fe粉和活性炭的质量之比。 (3)该“脱氧剂”的变质程度。 (1)1.2 g的固体是活性炭。 (1分) 8.0 g滤渣中Fe和FeO两种物质的总质量为8.0 g,1.2 g = 6.8 g (1分) 23112 (2)Fe元素的质量为:8.0 g= 5.6 g160则Fe粉和活性炭的质量之比为5.6?1.2 = 14?3 (1分) (3)由于铁变质而增加的氧元素的质量为:6.8 g,5.6 g = 1.2 g 2Fe 3O 112 48 X 1.2 48 (1分) 变质铁
29、的质量为1.2 g ?= 2.8 g 112ssing. After installation is complete if you have damaged galvanized parts shouldProce ve items, in order to ensure that the decoration construction progress and quality, trying to avoid construction does not .sealing process, required to support the outer decoration of a hole,
30、 hole, plugging holes and so on. This works more decoratiacing ase fixed curtain wall structure 16, decorative surfaces and decorative surface and joints between parts of other buildings fnts,) 15, and After the detailed design drawings issued a wall embedded part is inconsistent with the tenderer n
31、eeds to increation sewing (settlement sewing, and expansion joimine 12, and daughter wall top processing 13, and pan light ming, engineering of tie construction 14, and buildings of deform-urtain wall between of layer between, and around fire seal blocking 11, and antiand coutdoor advertising (conta
32、ining roofing and the State surface advertising) 9 to spring glass door 10, and concrete structure uried pieces 6, and light steel points type glass rain canopy, and aluminum Board rain canopy 7, and aluminum Venetian 8, andtain wall 4, and stone curtain wall 5, and curtain wall b2000 scope of this
33、project are: 1, 2, 3 aluminium sheet curtain wall of glass curtain wall and the Aluminum plastic Board cur-steel bolts, screws and studs GB/T3098.64 考试必备 2.8g (1分) 脱氧剂的变质程度=100% = 50% 5.6g20、在一个密闭容器中,有甲、乙、丙、丁四种物质在一定条件下充分反应后,测得反应前后各物质的质量如下表: 下列说法错误的是 物质 甲 乙 丙 丁 A该反应是分解反应 B甲可能是该反应反应前质量,5 2 20 22 的催化剂
34、g C乙、丙变化的质量比为9:8 D反应后甲的质量为反应后质量,待测 11 28 5 0g g 21、电解水时,常常要加入少量氢氧化钠使反应容易进行。 现将加有氢氧化钠的水通电一段时间后,产生lg氢气,其中氢氧化钠的质量分数也由4(8,变为5,。 计算:(1)生成氧气的质量。 (2)电解后剩余水的质量。 m(O2)=8g m(H2O)=205.2g 2H2O=2H2+O2 36 4 32 X 1 y x=9g y=8g 电解后剩余水的质量为m (m+9)*4.8%=m*5% m=205.2g 22、家里蒸馒头用的纯碱中含有少量的氯化钠,课外探究小组的同学欲测定纯碱中碳酸钠的含量。他们取该纯碱样
35、品11.0g,全部溶解在100.0g水中,再加入氯化钙溶液141.0g,恰好完全反应。过滤干燥后,称得沉淀质量为10.0g。请计算:(1)纯碱样品中碳酸钠的质量;(2)反应后所得滤液中溶质的质量分数。 设碳酸钠的质量为x,生成氯化钠的质量为y Na2CO3+CaCl2=CaCO3+2NaCl 106100117 x10y 解得x=10.6,y=11.7 因此,纯碱样品中碳酸钠的质量为10.6g 那么氯化钠的质量为0.4g,反应后溶液中一共含有氯化钠的质量为11.7+0.4=12.1g 反应后溶液的质量为:11g+100g+141g-10g=242g 那么氯化钠的质量分数为12.1/242*10
36、0%=5% 23、50g Ca(NO)溶液与50g KCO溶液混合后,恰好完全反应。经过滤、干燥、称量,得到5g3223沉淀。反应的化学方程式是:KCO+Ca(NO)=CaCO?+2KNO。请计算: 233233(1) 参加反应的KCO的质量。 23(2) 过滤后所得溶液的溶质质量分数。 七、学困生辅导和转化措施设K2CO3的质量为x,生成KNO3的质量为y 当a0时,抛物线开口向上,并且向上方无限伸展。当a0时,抛物线开口向下,并且向下方无限伸展。K2CO3 + Ca(NO3)2 = CaCO3? + 2KNO3 7、课堂上多设计一些力所能及的问题,让他们回答,并逐步提高要求。138。100
37、。202 zed parts shouldand quality, trying to avoid construction does not . Processing. After installation is complete if you have damaged galvaniress e, hole, plugging holes and so on. This works more decorative items, in order to ensure that the decoration construction progsurface and joints between
38、 parts of other buildings facing sealing process, required to support the outer decoration of a hol ativeedded part is inconsistent with the tenderer needs to increase fixed curtain wall structure 16, decorative surfaces and decorof deformation sewing (settlement sewing, and expansion joints,) 15, a
39、nd After the detailed design drawings issued a wall emb mine 12, and daughter wall top processing 13, and pan light ming, engineering of tie construction 14, and buildings-ng 11, and antiing) 9 to spring glass door 10, and concrete structure and curtain wall between of layer between, and around fire
40、 seal blockivertisd aluminum Board rain canopy 7, and aluminum Venetian 8, and outdoor advertising (containing roofing and the State surface adcurtain wall 4, and stone curtain wall 5, and curtain wall buried pieces 6, and light steel points type glass rain canopy, an2000 scope of this project are:
41、1, 2, 3 aluminium sheet curtain wall of glass curtain wall and the Aluminum plastic Board -steel bolts, screws and studs GB/T3098.65 x=6.9。 5g。y = 10.1 溶液的质量=50g + 50g5g = 95g K2CO3的质量为6.9g,溶液的溶质质量分数为10.6% (4)二次函数的图象:是以直线为对称轴,顶点坐标为(,)的抛物线。(开口方向和大小由a来决定)24、已知某石灰石样品中碳元素的质量分数为9%,则该石灰石样品中CaCO的质量分数是 3假设杂
42、质中不含碳元素,要不然就不能算了 根据碳和碳酸钙的相对分子质量比(就是质量比)就可以算出碳酸钙的质量分数了 C/CaCO=12/100 9%/(12/100)=75% 35.方位角:从某点的指北方向按顺时针转到目标方向的水平角,叫做方位角。如图3,OA、OB、OC的方位角分别为45、135、225。25、在一石灰石样品其中混有一些不含钙元素的杂质,经分析其中含有CaCO90%,,则样品中钙的3质量分数是多少 Ca占CaCO质量百分比为40/(40+12+3*16)=40% 3所以钙在石灰石中质量百分率为40%*90%=36% 面对新的社会要求,教师与学生应首先走了社会的前边,因此我们应该以新课
43、标要求为指挥棒,采用所有可行的措施,尽量体现以人为本,培养学生创新,开放的思维方式。另一方面注意处理好内容与思想的衔接,内容要在学生上学期的水平之上发展并为以后学习打下基础,思想上注意新思维与我国传统的教学思想结合26、甲醇(CHOH)是一种剧毒的,具有酒精气味的液体,现把一定量的甲醇与4.4克氧气混合于3一个密闭容器内,引燃发生如下反应:8CHOH,XO,mCO, nCO,16HO,当反应物完全转3222化为生成物后,降至常温常压下,容器内余下的液体为3.6克,求:(,)系数m与n的和;(2)该反应生成的CO的质量;(3)系数X的值(计算精确到0.01) 26 确定圆的条件:从8 CH3OH
44、 + x O2 = m CO2 + n CO + 16 H2O 可得: m + n = 8 (3分) 等弧:在同圆或等圆中,能够互相重合的弧叫做等弧。x = 11 (3分) (2)设生成CO2的质量为y( 8 CH3OH + 11O2 = 6 CO2 + 2 CO + 16 H2O 352 264 4.4克 y 3.确定二次函数的表达式:(待定系数法)352 :264 = 4(4克 :y (3分) y =3.30克 (3分) 27、将氢气通入装有m克氧化铜的试管中,加热一段时间后停止加热,有部分氧化铜被还原,此1时,试管内固体物质的质量变为m克,(其中m?m)试求:m克固体物质中含有多少克金属
45、铜, 2122推论1: 同弧或等弧所对的圆周角相等。28、取一定的氯酸钾和二氧化锰的混合物,其中二氧化锰质量分数为20%,加热一段时间后,二氧化锰含量提高到25%,求此时氯酸钾的分解率。 29、用加热氯酸钾和二氧化锰混合物和方法制取氧气,已知二氧化锰的质量为0.3克,当制得所需氧气后,冷却,称量剩余的固体物质,其质量为10克,继续加热,充分分解,最后又制得氧气1.92克。求开始时取用的氯酸钾和二氧化锰混合物中氯酸钾的质量分数, ssing. After installation is complete if you have damaged galvanized parts shouldPro
46、ce ve items, in order to ensure that the decoration construction progress and quality, trying to avoid construction does not .sealing process, required to support the outer decoration of a hole, hole, plugging holes and so on. This works more decoratiacing ase fixed curtain wall structure 16, decora
47、tive surfaces and decorative surface and joints between parts of other buildings fnts,) 15, and After the detailed design drawings issued a wall embedded part is inconsistent with the tenderer needs to increation sewing (settlement sewing, and expansion joimine 12, and daughter wall top processing 13, and pan light ming, engineering of tie construction 14, and buildings of deform-urtain wall between of layer between, and around fire seal blocking 11, and antiand coutdoor advertising (containing roofing and the State surface advertising) 9 to spring glass door 10, and concrete
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