$\begin{array}{l} CuO + {H_2}S{O_4} \to CuS{O_4} + {H_2}O\\ nCuO = \dfrac{{16}}{{80}} = 0,2\,mol\\ n{H_2}S{O_4} = nCuO = 0,2\,mol\\ m{\rm{dd}}{H_2}S{O_4} = \dfrac{{0,2 \times 98}}{{20\% }} = 98g\\ m{H_2}O\,bd = 98 - 0,2 \times 98 = 78,4g\\ m{H_2}O\,spu = 78,4 + 0,2 \times 18 = 82g\\ mCuS{O_4}.{H_2}O = a\,g\\ mCuS{O_4}\,kt = 0,64a\\ mCuS{O_4}\,bd = 0,2 \times 160 - 32g\\ mCuS{O_4}\, = 32 - 0,64a\,g\\ m{H_2}O\,kt = 0,36a\,g\\ m{H_2}O = 82 - 0,36a\,g\\ \dfrac{{32 - 0,64a}}{{82 - 0,36a}} = \dfrac{{17,4}}{{100}}\\  =  > a = 30,71g \end{array}$

`#namth`

Đáp án:

Giải thích các bước giải:

nCuO=1680=0,2molnCuO=1680=0,2mol

PT:CuOCuO+H2SO4→CuSO4+H2O.H2SO4→CuSO4+H2O.

Theo pt, ta có: nCuO=nCuSO4=nH2SO4=0,2molnCuO=nCuSO4=nH2SO4=0,2mol

⇒mCuSO4=0,2.160=32gam,mH2SO4=0,2.98=19,6gam.⇒mCuSO4=0,2.160=32gam,mH2SO4=0,2.98=19,6gam.

⇒mddH2SO4=29,6:20%=98gam.⇒mddH2SO4=29,6:20%=98gam.

⇒mddsauphảnứng=98+16=114gam⇒mH2O=114−32=82gam⇒mddsauphảnứng=98+16=114gam⇒mH2O=114−32=82gam

Gọi nCuSO4.5H2OnCuSO4.5H2O là a mol.

nCuSO4=nCuSO44.5H2O=amol→mCuSO4=160agam.nCuSO4=nCuSO44.5H2O=amol→mCuSO4=160agam.

nnH2O=5.nCuSO4.5H2O=5agam→mH2O=90agam.nH2O=5.nCuSO4.5H2O=5agam→mH2O=90agam.

Ta có pt:32−160a82−90a.100=17,4⇒a=0,123mol32−160a82−90a.100=17,4⇒a=0,123mol→mCuSO4.5H2O=0,123.250=30,7gam.→mCuSO4.5H2O=0,123.250=30,7gam.

Vậy khối lượng tinh thể CuSO4.5H2OCuSO4.5H2O đã tách ra khỏi dd là 30,7 gam