\(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(CuO+2HCl\rightarrow CuCl_2+H_2O\)
a, Ta có:
\(n_{H2}=\frac{0,224}{22,4}=0,01\left(mol\right)\)
\(\Rightarrow n_{Mg}=0,01\left(mol\right)\)
\(\%m_{Mg}=\frac{0,01.24}{1,04}.100\%=23,08\%\)
\(\%m_{CuO}=100\%-23,08\%=76,92\%\)
b,\(n_{CuO}=\frac{0,8}{80}=0,01\left(mol\right)\)
\(\Rightarrow n_{HCl}=0,01.2+0,01.2=0,04\)
\(CM_{HCl}=\frac{0,04}{0,25}=0,16M\)
\(\Rightarrow m=m_{MgCl2}+m_{CuCl2}=0,01.95+0,01.135=2,3\left(g\right)\)