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 \(n_{CuO}=\dfrac{1,6}{80}=0,02\left(mol\right)\)
\(=>Cm_{H_2SO_4}=\dfrac{0,02}{0,1}=0,2\left(M\right)\)