问题 问答题

将足量铁投入到200mLH2SO4和CuSO4的混和溶液中,充分反应后,产生H21.12L(标准状况),铁块的质量减轻了1.2g.求原溶液中c(H2SO4)及c(CuSO4).(假设反应前后溶液的体积不变)

答案

Fe+H2SO4=FeSO4 +H2

     56g  1mol    22.4L

 m1(Fe) n(H2SO4) 1.12L

m1(Fe)=

56g×1.12L
22.4L
=2.8g

n(H2SO4)=

1mol×1.12L
2.24L
=0.05mol

V(液)=V(H2SO4)=V(CuSO4)=200mL=0.2L

c(H2SO4)=

n(H2SO4)
V(H2SO4)
=
0.05mol
0.2L
=0.25mol/L

     Fe+CuSO4=FeSO4+Cu

    56g  1mol      64g

m2(Fe) n(CuSO4)  m(Cu)

m(Cu)=

64g
56g
×m2(Fe)=
8
7
m2(Fe)

由题意得:m1(Fe)+m2(Fe)-m(Cu)=1.2g

2.8g+m2(Fe)-

8
7
m2(Fe)=1.2g

m2(Fe)=11.2g

n(CuSO4)=

1mol×m2(Fe)
56g
=
1mol×11.2g
56g
=0.2mol

c(CuSO4)=

n(CuSO4)
V(CuSO4)
=
0.2mol
0.2L
=1mol/L

答:原溶液中c(H2SO4)=0.25mol/L,c(CuSO4)=1mol/L.

阅读理解
阅读理解。
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