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课时跟踪检测(四) 物质的量在方程式计算中的应用

课时跟踪检测(四)  物质的量在方程式计算中的应用
课时跟踪检测(四)  物质的量在方程式计算中的应用

课时跟踪检测(四) 物质的量在方程式计算中的应用

1.在反应X +2Y===R +2M 中,已知R 和M 的摩尔质量之比为22∶9,当1.6 g X 与Y 完全反应后,生成4.4 g R 。则在此反应中Y 和M 的质量之比为( ) A .16∶9 B .23∶9 C .32∶9

D .46∶9

解析:选A 由已知反应:

X +2Y===R + 2M

M r (R) 2M r (M) 4.4 g m

由题意得M r (R)∶2M r (M)=22∶18=4.4 g ∶m ,解得m =3.6 g 。

根据质量守恒定律,参加反应的Y 的质量为4.4 g +3.6 g -1.6 g =6.4 g ,所以Y 与M 的质量比为6.4 g ∶3.6 g =16∶9。

2.(2020·吉林联考)把a L 含硫酸铵和硝酸铵的混合溶液分成两等份。一份加入含b mol NaOH 的溶液并加热,恰好把NH 3全部赶出;另一份需消耗c mol BaCl 2才能使SO 2-

4完全沉淀,则原溶液中NO -

3的物质的量浓度(mol·L -

1)为( ) A.b -2c a

B .2b -4c a

C.2b -c a

D .b -4c a

解析:选B n (NH +4)=b mol ,n (SO 2-4)=c mol ,根据电荷守恒知n (NO -3

)=(b -2c )mol ,c (NO -3)=

b -2c

a 2

mol·L -1。 3.在Al 2(SO 4)3、K 2SO 4和明矾的混合溶液中,如果c (SO 2-

4)等于0.2 mol·L -

1,当加入等体积的0.2 mol·L -1

的KOH 溶液时,生成的沉淀恰好溶解,则原混合溶液中K +

的物质的量

浓度为( ) A .0.25 mol·L -1 B .0.2 mol·L -

1 C .0.45 mol·L -1

D .0.225 mol·L -

1

解析:选A 根据Al 3++4OH -===AlO -2+2H 2O 可知,加入等体积的KOH 溶液时生成的沉淀恰好溶解,说明原溶液中c (Al 3+)=1

4

×0.2 mol·L -1=0.05 mol·L -1。设K +的物质

的量浓度为x mol·L -1,则根据电荷守恒可知,c (K +)+c (Al 3+)×3=c (SO 2-4)×2,即x

mol·L -1+0.05 mol·L -1×3=0.2 mol·L -1×2,解得x =0.25。

4.工业上,利用黄铜矿(主要成分是CuFeS 2)冶炼金属,产生的废气可以制备硫酸。某黄铜矿中铜元素的质量分数为a %(假设杂质不含铜、铁、硫元素),其煅烧过程转化率为75%,得到的SO 2转化为SO 3的转化率为80%,SO 3的吸收率为96%。现有黄铜矿100 t ,其废气最多能制备98%的硫酸( ) A .1.8a t B .2.8a t C .3.2a t

D .4.5a t

解析:选A 根据题中转化过程中物质变化及物质中硫原子守恒可得关系式:CuFeS 2~2SO 2~2SO 3~2H 2SO 4,根据黄铜矿中铜元素的质量分数为a %可得100 t 黄铜矿中n (CuFeS 2)=n (Cu)=100×106 g ×a %64 g·mol -1=1×106×a 64 mol ,则根据关系式及各步的转化率可

知n (H 2SO 4)=2n (CuFeS 2)×75%×80%×96%=1.8a ×104 mol ,则能制备98%的硫酸的质量为1.8a ×104 mol ×98 g·mol -1

98%

=1.8a ×106 g =1.8a t 。

5.乙烯和乙烷的混合气体共a mol ,与b mol O 2共存于一密闭容器中,点燃后充分反应,乙烯和乙烷全部消耗完,得到CO 和CO 2的混合气体和45 g H 2O ,试求: (1)当a =1时,乙烯和乙烷的物质的量之比n (C 2H 4)∶n (C 2H 6)=________。

(2)当a =1,且反应后CO 和CO 2混合气体的物质的量为反应前O 2的2

3时,b =________,

得到的CO 和CO 2的物质的量之比n (CO)∶n (CO 2)=________。

解析:(1)设原混合气体中C 2H 4、C 2H 6的物质的量分别为x 、y ,则

?

??

x +y =1 mol 4x +6y =45 g

18 g·mol -

1

×2 解得x =y =0.5 mol ,所以n (C 2H 4)∶n (C 2H 6)=1∶1。

(2)根据碳原子守恒,n (CO)+n (CO 2)=2[n (C 2H 4)+n (C 2H 6)]=2a mol ,因为a =1,且n (CO)+n (CO 2)=2b 3 mol ,所以2=2b

3,b =3,设反应后混合气体中CO 、CO 2的物质的量分别

为x 、y ,则:

?

??

x +y =2 mol (碳原子守恒)

x +2y =3 mol ×2-45 g

18 g·mol -

1

(氧原子守恒)

解得x =0.5 mol ,y =1.5 mol ,所以n (CO)∶n (CO 2)=1∶3。

答案:(1)1∶1 (2)3 1∶3

6.水泥是重要的建筑材料。水泥熟料的主要成分为CaO 、SiO 2,并含有一定量的铁、铝和镁等金属的氧化物。实验室测定水泥样品中钙含量的过程如图所示:

草酸钙沉淀经稀H 2SO 4处理后,用KMnO 4标准溶液滴定,通过测定草酸的量可间接获知钙的含量,滴定反应为MnO -

4+H +

+H 2C 2O 4―→Mn 2+

+CO 2+H 2O 。实验中称取0.400 g 水泥样品,滴定时消耗了0.050 0 mol·L -1

的KMnO 4溶液36.00 mL ,则该水泥样品中钙的质

量分数为________。

解析:根据反应中转移电子数相等可找出关系式2MnO -4~5H 2C 2O 4,结合消耗KMnO 4溶液的浓度和体积可求出n (H 2C 2O 4)=0.050 0 mol·L -1×36.00×10-3 L ×52=4.5×10-3 mol ,则该水

泥样品中钙的质量分数为4.5×10-3 mol ×40 g·mol -1

0.400 g ×100%=45.0%。

答案:45.0%

7.过氧化钙晶体(CaO 2·8H 2O)可用于改善地表水质、处理含重金属粒子废水、应急供氧等。

(1)已知:I 2+2S 2O 2-

3===2I -

+S 4O 2

6,测定制备的过氧化钙晶体中CaO 2的含量的实验步

骤如下:

第一步:准确称取a g 产品放入锥形瓶中,再加入过量的b g KI 晶体,加入适量蒸馏水溶解,再滴入少量2 mol·L -1

的H 2SO 4溶液,充分反应。

第二步:向上述锥形瓶中加入几滴淀粉溶液。

第三步:逐滴加入浓度为c mol·L

-1

的Na 2S 2O 3溶液发生反应,滴定达到终点时出现的现

象是________________________________________________________________________ ___________________________________________________________________________。 若滴定消耗Na 2S 2O 3溶液V mL ,则样品中CaO 2的质量分数为________(用字母表示)。

(2)已知过氧化钙加热至350 ℃左右开始分解放出氧气。将过氧化钙晶体(CaO 2·8H 2O)在坩埚中加热逐渐升高温度,测得样品质量随温度的变化如图所示,则350 ℃左右所得固体物质的化学式为______________。

解析:(1)根据得失电子守恒,可得关系式CaO 2~I 2~2S 2O 2-3,则样品中CaO 2

的质量分数为cV ×10-3×722×a ×100%=36cV ×10-3

a ×100%。

(2)CaO 2·8H 2O 的摩尔质量为216 g·mol -1,故2.16 g 过氧化钙晶体为0.01 mol,350 ℃左右所得固体质量为0.56 g ,根据钙原子守恒,可知为CaO 。

答案:(1)溶液由蓝色变无色,且30 s 内不恢复蓝色 36cV ×10-

3

a

×100% (2)CaO

8.人体血液里Ca 2+

的浓度一般采用mg·cm -3

来表示。抽取一定体积的血样,加适量的草酸

铵[(NH 4)2C 2O 4]溶液,可析出草酸钙(CaC 2O 4)沉淀,将此草酸钙沉淀洗涤后溶于强酸可得草酸(H 2C 2O 4),再用酸性KMnO 4溶液滴定即可测定血液样品中Ca 2+

的浓度。

抽取血样20.00 mL ,经过上述处理后得到草酸,再用0.020 mol·L

-1酸性KMnO 4溶液滴

定,使草酸转化成CO 2逸出,这时共消耗12.00 mL 酸性KMnO 4溶液。

(1)已知草酸与酸性KMnO 4溶液反应的离子方程式为2MnO -

4+5H 2C 2O 4+6H +

===2Mn x +

+10CO 2↑+8H 2O ,则其中的x =________。 (2)经过计算,血液样品中Ca 2+

的浓度为______mg·cm -

3。

解析:(1)由电荷守恒可得x =2。(2)血样处理过程中发生反应的离子方程式依次是:①

Ca 2++C 2O 2-4===CaC 2O 4↓;②CaC 2O 4+2H +===Ca 2++H 2C 2O 4;③2MnO -4+5H 2C 2O 4

6H +===2Mn 2++10CO 2↑+8H 2O ,由此可得关系式:5Ca 2+~5CaC 2O 4~5H 2C 2O 4~2MnO -4,所以n (Ca 2+)=52n (MnO -4

)=52×0.012 00 L ×0.020 mol·L -1=6.0×10-4 mol ,血液样品中Ca 2+的浓度=40 g·mol -1×6.0×10-4 mol

20.00 cm 3=1.2×10-3 g·cm -3=1.2 mg·cm -3。

答案:(1)2 (2)1.2

9.利用钴渣[含Co(OH)3、Fe(OH)3等]制备钴氧化物的工艺流程如下:

新人教版必修一 Unit5 Period 1课时跟踪检测

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