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

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

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

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

:)

nCuO=16/80=0,2(mol)

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

0,2___________0,2_____0,2(mol)

b) mCuSO4=160.0,2=32(g)

c) mH2SO4=0,2.98=19,6(g)

=>C%ddH2SO4= (19,6/100).100=19,6%

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

Bài này anh có hỗ trợ ở dưới rồi nha em!

nCuO=16/80=0,2(mol)

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

0,2___________0,2_____0,2(mol)

b) mCuSO4=160.0,2=32(g)

c) KL dung dịch H2SO4 tham gia p.ứ là đề bài cho 100(g)

d) mH2SO4=0,2.98=19,6(g)

=>C%ddH2SO4= (19,6/100).100=19,6%

a, \(2HCl+Ba\left(OH\right)_2\rightarrow BaCl_2+2H_2O\)

b, \(n_{HCl}=0,06.0,1=0,006\left(mol\right)\)

Theo PT: \(n_{Ba\left(OH\right)_2}=\dfrac{1}{2}n_{HCl}=0,003\left(mol\right)\)

\(\Rightarrow V_{Ba\left(OH\right)_2}=\dfrac{0,003}{0,2}=0,015\left(l\right)=15\left(ml\right)\)

c, \(n_{BaCl_2}=\dfrac{1}{2}n_{Ba\left(OH\right)_2}=0,003\left(mol\right)\Rightarrow C_{M_{BaCl_2}}=\dfrac{0,003}{0,06+0,015}=0,04\left(M\right)\)

\(a/2HCl+Ba\left(OH\right)_2\rightarrow BaCl_2+2H_2O\\ b/n_{HCl}=0,06.0,1=0,006mol\\ n_{Ba\left(OH\right)_2}=n_{BaCl_2}=0,006:2=0,003mol\\ V_{Ba\left(OH\right)_2}=\dfrac{0,003}{0,2}=0,015l\\ c/C_{M_{BaCl_2}}=\dfrac{0,003}{0,06+0,015}=0,04M\)