a. \(PTHH:4X+3O_2--->2X_2O_3\left(1\right)\)
Ta có: \(m_X=m_{X_{\left(X_2O_3\right)}}=8,1\left(g\right)\)
\(\Rightarrow m_{O_{\left(X_2O_3\right)}}=15,3-8,1=7,2\left(g\right)\)
Ta lại có: \(m_{O_2}=m_{O_{\left(X_2O_3\right)}}=7,2\left(g\right)\)
\(\Rightarrow n_{O_2}=\dfrac{7,2}{32}=0,225\left(mol\right)\)
Theo PT(1): \(n_X=\dfrac{4}{3}.n_{O_2}=\dfrac{4}{3}.0,225=0,3\left(mol\right)\)
\(\Rightarrow M_X=\dfrac{8,1}{0,3}=27\left(g\right)\)
Vậy X là nguyên tố nhôm (Al)
b. PTHH:
\(Al+2HCl--->AlCl_3+H_2\left(2\right)\)
\(CuO+2HCl--->CuCl_2+H_2O\left(3\right)\)
Ta có: \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
- Theo PT(2): \(n_{Al}=n_{H_2}=0,3\left(mol\right)\)
\(\Rightarrow m_{Al}=0,3.27=8,1\left(g\right)\)
\(\Rightarrow m_{CuO}=13,4-8,1=5,3\left(g\right)\)
- Ta có:
\(m_{dd_{sau.phản.ứng}}=m_{hh}+m_{dd_{HCl}}-m_{H_2}=13,4+200-0,3.2=212,8\left(g\right)\)
Ta có: \(n_{CuO}=\dfrac{5,3}{80}=0,06625\left(mol\right)\)
Theo PT(2): \(n_{AlCl_3}=n_{Al}=0,3\left(mol\right)\)
\(\Rightarrow m_{AlCl_3}=0,3.133,5=40,05\left(g\right)\)
\(\Rightarrow C_{\%_{AlCl_3}}=\dfrac{40,05}{212,8}.100\%=18,82\%\)
Theo PT(3): \(n_{CuCl_2}=n_{CuO}=0,06625\left(mol\right)\)
\(\Rightarrow m_{CuCl_2}=0,06625.135=8,94375\left(g\right)\)
\(\Rightarrow C_{\%_{CuCl_2}}=\dfrac{8,94375}{212,8}.100\%=4,2\%\)
