\(CuO+H2SO4-->CuSO4+H2O\)
\(n_{CuO}=\frac{40}{80}=0,5\left(mol\right)\)
\(n_{H2SO4}=n_{C_{ }uO}=0,5\left(mol\right)\)
\(m_{H2SO4}=0,5.98=49\left(g\right)\)
\(m_{ddH2SO4}=\frac{49.100}{9,8}=500\left(g\right)\)
m dd sau pư= 500+ 40=540(g)
\(n_{CuSO4}=n_{CuO}=0,5\left(mol\right)\)
\(m_{CuSO4}=0,5.160=80\left(g\right)\)
\(C\%CuSO4=\frac{80}{540}.100\%=14,81\%\)