Dip an iron bar weighing 100g into 500ml of mixed solution of CuSO4 0.08M and Ag2SO4 0.004M. After a while, take the iron bar and weigh it again, the mass is 100.48 g, calculate the mass of metal attached to the iron bar and the concentration molarity of substances in the reaction solution Assume the solution volume remains unchanged

$\Delta m=100，48-100=0，48g$

$n_{CuSO_4}=0，5.0，08=0，04(mol)$

$n_{Ag_2SO_4}=0，5.0，004=0，002(mol)$

Thứ tự phản ứng$:Ag_2SO_4， CuSO_4$

$*$ Nếu $Ag_2SO_4$ vừa hết $:$

$Fe+Ag_2SO_4\to FeSO_4+2Ag$ $(1)$

$n_{Fe\text{ws}}=0，002(mol); n_{Ag}=0，004(mol)$

$\Rightarrow \Delta m=0，004.108-0，002.56=0，32g$

$*$ Nếu $CuSO_4$ vừa hết $:$

$Fe+CuSO_4\to FeSO_4+Cu$ $(2)$

$n_{Fe\text{pứ}}=n_{Cu}=0，04(mol)$

$\Rightarrow \Delta m=0，04.64+0，004.108-56(0，04+0，002)=0，64g$

Vì $0，32<0，48<0，64$ nên $Ag_2SO_4$ hết $,$ $CuSO_4$ phản ứng $1$ phần.

Sau khi $(1)$ kết thúc $:$

$n_{Fe\text{ws}}=n_{FeSO_4}=0，002(mol)$

$n_{Ag}=0，004(mol)$

Gọi $a$ là số mol $CuSO_4$ phản ứng .

$\Rightarrow n_{Fe\text{ws}}=n_{CuSO_4}=n_{Cu}=x (mol)$

$0，004.108+64x-56(0，002+x)=0，48$

$\Leftrightarrow x=0，02$

$\Rightarrow n_{Cu}=0，02(mol)$

$\to m_{\text{kim loại bám}}=0，02.64+0，004.108=1，712g$

Dư $0，04-0，02=0，02$ mol $CuSO_4$

$\Rightarrow C_{M_{CuSO_4}}=0，04M$

$n_{FeSO_4\text{tổng}}=n_{Fe\text{ws}\text{tổng}}=0，02+0，002=0，022(mol)$

$\Rightarrow C_{M_{FeSO_4}}=0，044M$