\(18,\\ n_{FeO}=\dfrac{7,2}{72}=0,1mol\\ FeO+H_2SO_4\rightarrow FeSO_4+H_2O\\ m_{ddH_2SO_4}=100.1=100g\\ n_{H_2SO_4}=\dfrac{100.10,95}{100}:36,5=0,3mol\\ \Rightarrow\dfrac{0,1}{1}< \dfrac{0,3}{1}\Rightarrow H_2SO_4.dư\\ n_{FeSO_4}=n_{H_2SO_4}=n_{FeO}=0,1mol\\ C_{\%FeSO_4}=\dfrac{0,1.152}{100+7,2}\cdot100=14,18\%\\C_{\%H_2SO_4}=\dfrac{\left(0,3-0,1\right)98}{100+7,2}\cdot100=18,28\%\)
\(18.\\ CuO+H_2SO_{4\left(loãng\right)}\rightarrow CuSO_4+H_2O\\ Cu+2H_2SO_{4\left(đặc,t^0\right)}\rightarrow CuSO_4+2H_2O+SO_2\)
\(20.\\ n_{CuO}=\dfrac{12}{80}=0,15mol\\ n_{HCl}=0,4.1=0,4mol\\CuO+2HCl\rightarrow CuCl_2+H_2O\\ \Rightarrow\dfrac{0,15}{1}< \dfrac{0,4}{2}\Rightarrow HCl,dư\\ n_{CuCl_2}=n_{CuO}=0,15mol\\C_{M_{CuCl_2}}=\dfrac{0,15}{0,4}=0,375M\\ C_{M_{HCl\left(dư\right)}}=\dfrac{0,4-0,15.2}{0,4}=0,25M\)