nCuO=3,2:80=0,04 mol

PTHH: CuO+H2SO4=>CuSO4+H2O

       0,04mol->0,04mol->0,04mol->0,04mol

=>  m H2SO4=0,04.98=3,92g

=> m ddH2SO4 tham gia phản ứng =\(\frac{3,92.100}{4,9}=80\)g

 theo địnhluật bảo toàn khối lượng => m CuSO4= mCuO+mH2SO4-mH2O=3,2+80-0,04.18=82,48g

m CuSO4 thu được= 0,04.160=6,4g

=> C% CuSO4 =\(\frac{6,4}{82,48}.100=7,76\%\)

:)

Mình nghĩ là vậy.. 

a. \(pthh:CuO+H_2SO_4--->CuSO_4+H_2O\)

b. Ta có: \(n_{CuO}=\dfrac{3,2}{80}=0,04\left(mol\right)\)

Theo pt: \(n_{CuSO_4}=n_{CuO}=0,04\left(mol\right)\)

\(\Rightarrow m_{CuSO_4}=0,04.160=6,4\left(g\right)\)

\(m_{dd_{CuSO_4}}=3,2+80=83,2\left(g\right)\)

\(\Rightarrow C_{\%_{CuSO_4}}=\dfrac{6,4}{83,2}.100\%=7,69\%\)

`a)PTHH:`

`CuO + H_2 SO_4 -> CuSO_4 + H_2 O`

`0,04`        `0,04`                  `0,04`      `0,04`    `(mol)`

`n_[CuO]=[3,2]/80=0,04(mol)`

`b)C%_[CuSO_4]=[0,04.160]/[3,2+80-0,04.2].100~~7,7%`

(Đề bài thiếu dữ kiện để tính khối lượng dung dịch)

$CuO + H_2SO_4 \to CuSO_4 + H_2O$
$n_{CuSO_4} = n_{CuO} = \dfrac{8}{80} = 0,1(mol)$
$m_{CuSO_4} = 0,1.160 = 16(gam)$

Sau phản ứng : 

$m_{dd} = m_{CuO} + m_{dd\ H_2SO_4} = 8 + m_{dd\ H_2SO_4}(gam)$
Suy ra : 

$C\%_{CuSO_4} = \dfrac{16}{8 + m_{dd\ H_2SO_4}}.100\%$

CuO   +    H₂SO₄ →   CuSO₄   +   H₂O

0,2...............0,2.............0,2..............................................(mol)

\(m_{dd\ H_2SO_4} = \dfrac{0,2.98}{20\%} = 98(gam)\\ \Rightarrow m_{dd\ sau\ pư} = 98 + 0,2.80 = 114(gam)\\ m_{CuSO_4} = 0,2.160 = 32(gam)\\ \Rightarrow m_{H_2O} = 114 - 32 = 82(gam)\)

Gọi \(n_{CuSO_4.5H_2O} = a(mol)\).

Sau khi tách tinh thể, dung dịch còn :

\(m_{CuSO_4} = 32 - 160a(gam)\\ m_{H_2O} = 82 - 18.5a = 82 - 90a(gam)\)

Suy ra:

 \(\dfrac{32-160a}{82-90a} =\dfrac{17,4}{100}\\ \Rightarrow a = 0,12284\\ \Rightarrow m_{CuSO_4.5H_2O} = 0,12284.250 = 30,71(gam)\)

a=0,12284 kiểu gì vậy

Với 250 từ đâu ra vậy

CuO+H₂SO₄->CuSO₄+H₂O

0,02----0,02 mol

n _CuO=\(\dfrac{1,6}{80}\)0,02 mol

=>Cm _H2SO4=\(\dfrac{0,02}{0,1}\)=0,2 M

Đổi 100 ml = 0,1 l

CuO + H₂SO₄ -> H₂O + CuSO₄

Ta có \(n_{CuO}=\dfrac{1,6}{80}=0,02\left(mol\right)\)

\(=>Cm_{H_2SO_4}=\dfrac{0,02}{0,1}=0,2\left(M\right)\)

\(nCuO=\dfrac{8}{80}=0,1\left(mol\right)\)

\(nH_2SO_4=\dfrac{19,6}{98}=0,2\left(mol\right)\)

\(LTL:\dfrac{0,1}{1}< \dfrac{0,2}{1}\)

=> CuO pứ đủ , H2SO4 dư

CuO+H₂SO₄ -> CuSO₄+H₂O

0,1      0,1          0,1         0,1

\(nH_2SO_{4\left(dư\right)}=0,2-0,1=0,1\left(mol\right)\)

\(mH_2SO_{4\left(dư\right)}=0,1.98=9,8\left(g\right)\)

c1:

\(m\left(muối\right)=mCuSO_4=0,1.160=16\left(g\right)\)

c2:

\(mH_2O=0,1.18=1,8\left(g\right)\)

BTKL:

mCuO+mH₂SO₄   = m CuSO₄+ mH₂O

8          +  (19,6-9,8)  = m CuSO4  + 1, 8

=> mCuSO4 = 8 + ( 19,6 - 9,8 ) - 1,8 = 16 (g)

PTHH: CuO + H₂SO₄ --> CuSO₄ + H₂O

             0,2---->0,2-------->0,2---->0,2

=> \(m_{H_2SO_4}=0,2.98=19,6\left(g\right)\Rightarrow m_{dd.H_2SO_4}=\dfrac{19,6.100}{20}=98\left(g\right)\)

\(m_{H_2O\left(bđ\right)}=98-19,6=78,4\left(g\right)\)

Gọi số mol CuSO₄.5H₂O tách ra là a (mol)

\(n_{CuSO_4\left(tách.ra\right)}=a\left(mol\right)\) => \(n_{CuSO_4\left(dd.sau.khi.làm.nguội\right)}=0,2-a\left(mol\right)\)

\(n_{H_2O\left(tách.ra\right)}=5a\left(mol\right)\Rightarrow m_{H_2O\left(dd.sau.khi.làm.nguội\right)}=78,4+0,2.18-18.5a=82-90a\left(g\right)\)

Xét \(S_{10^oC}=\dfrac{160\left(0,2-a\right)}{82-90a}.100=17,4\left(g\right)\)

=> a = \(\dfrac{4433}{36085}\left(mol\right)\) => \(m_{CuSO_4.5H_2O}=\dfrac{4433}{36085}.250=30,7122\left(g\right)\)

\(a,n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\)

PTHH: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)

           0,05---->0,1------>0,1

\(\rightarrow x=C\%_{HCl}=\dfrac{0,1.36,5}{200}.100\%=1,825\%\)

\(b,\) PTHH: \(CuCl_2+2NaOH\rightarrow Cu\left(OH\right)_2\downarrow+2NaCl\)

                 0,05----->0,1-------->0,05----------->0,1

\(\rightarrow m_{ddNaOH}=\dfrac{0,1.40}{10\%}=40\left(g\right)\\ \rightarrow m_{dd\left(sau.pư\right)}=40+200+4-0,05.98=239,1\left(g\right)\)

\(\rightarrow C\%_{NaCl}=\dfrac{0,1.58,5}{239,1}.100\%=2,45\%\)

\(n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\\ pthh:CuO+HCl\rightarrow CuCl_2+H_2O\)
          0,05     0,05      0,05        0,05 
\(x=C\%_{HCl}=\dfrac{0,05.36,5}{200}.100\%=0,9125\%\\ pthh:CuCl_2+2NaOH\rightarrow Cu\left(OH\right)_2+2NaCl\)