nCuO=0,04 mol

CuO        + H2SO4 =>CuSO4 + H2O

0,04 mol=>0,04 mol=>0,04 mol

mH2SO4=0,04.98=3,92 gam

=>m dd H2SO4=3,92/4,9%=80 gam

mCuSO4 sau=0,04.160=6,4 gam

mdd CuSO4=3,2+80=83,2 gam

C% dd CuSO4=6,4/83,2.100%=7,69%

_{cho \(m_{CuO}=3,2g\Rightarrow n_{CuO}=\frac{3,2}{80}=0,04mol\)}

PTHH: 

  CuO           +          H₂SO₄       ->    CuSO₄    +   H₂O

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

ta có: \(m_{H_2SO_4}=0,04.98=3,92g\)

\(C\%_{d^2H_2SO_{4_{ }}}=4,9\%\)

=. \(m_{d^2H_2SO_4}=\frac{m_{H_2SO_4}.100}{C\%}=\frac{3,92.100}{4,9}=80g\) 

áp dụng ĐLBTKL ta có: \(m_{d^2CUSO_4}=m_{CuO}+m_{d^2H_2SO_4}=3,2+80=83,2g\)

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

\(\Rightarrow C\%_{d^2CuSO_4}=\frac{m_{CuSO_4}}{m_{d^2CuSO_4}}.100=\frac{6,4}{83,2}.100=7,69\%\)

a) \(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

b) \(n_{CuO}=\dfrac{3,2}{80}=0,04\left(mol\right)=n_{H_2SO_4}=n_{CuSO_4}\)

\(m_{ddH_2SO_4}=\dfrac{0,04.98}{4,9\%}=80\%\)

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

=> \(C\%_{CuSO_4}=\dfrac{0,04.160}{83,2}.100=7,69\%\)

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

:)

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\\ n_{H_2SO_4}=n_{CuO}=\dfrac{19,6}{98}=0,2\left(mol\right)\\ \Rightarrow a=m_{CuO}=0,2\cdot80=16\left(g\right)\)

n H2SO4 = \(\dfrac{19,6}{2+32+16.4}=0,2mol\)

\(CuO+H_2SO_4->CuSO_4+H_2O\)

0,2..........0,2

m CuO = 0,2.(64+16)=16 g 

Vậy a =16 

a)CuO+H2SO4--->CuSO4+H2O

b) Ta có

n CuO=3,2/80=0,04(mol)

Theo pthh

n H2SO4=n CuO=0,04(mol)

m dd H2SO4=\(\frac{0,04.98.100}{4,9}=80\left(g\right)\)

m dd sau pư=m CuO+m CuSO4=80+3,2=83,2(g)

n CuSO4=0,04(mol)

C% CuSO4=\(\frac{0,04.160}{83,2}.100\%=7,7\%\)

Mình nghĩ là vậy.. 

\(n_{CuO}=\dfrac{3,2}{80}=0,04\left(mol\right)\\ pthh:CuO+H_2SO_4\rightarrow CuSO_4+H_2O\) 
           0,04       0,04           0,04      0,04 
\(m_{\text{dd}_{H_2SO_4}}=\dfrac{\left(0,04.36,5\right).100}{4,9}=29,79\left(g\right)\\ m_{\text{dd}_{CuSO_4}}=3,2+29,79-\left(0,04.2\right)=32,91\left(g\right)\\ C\%_{\text{dd}}=\dfrac{0,04.160}{32,91}.100\%=19,44\%\)