$Mg+2HCl\rightarrow MgCl_2+H_2\\ Fe_2O_3+6HCl\rightarrow 2FeCl_3+3H_2O\\ 4.1\\  n_{Mg}=n_{H_2}=\dfrac{3,36}{22,4}=0,15(mol)\\ \%m_{Mg}=\dfrac{0,15.24}{19,6}.100\%=18,37\%\\ \%m_{Fe_2O_3}=100\%-18,37\%=81,63\%\\ m_{Mg}=0,15.24=3,6(g)\\ m_{Fe_2O_3}=19,6-3,6=16(g)\\ 4.2\\ n_{Fe_2O_3}=\dfrac{16}{160}=0,1(mol)\\  n_{HCl}=2n_{Mg}+6n_{Fe_2O_3}=0,9(mol)\\ m=m_{dd\ HCl}=\dfrac{0,9.36,5}{15\%}=219(g)\\ n_{MgCl_2}=n_{Mg}=0,15(mol)\\ n_{FeCl_3}=2n_{Fe_2O_3}=0,2(mol)\\ m_1=m_{MgCl_2}+m_{FeCl_3}\\ =0,15. 95+0,2.162,5=46,75(g)$

 ` Mg + 2HCl -> MgCl_2 + H_2 `

` Fe_2O_3 + 6HCl -> 2FeCl_3 + 3H_2O ` 

` n_(H_2) = ( 3,36 ) / ( 22,4 ) = 0,15 ( mol ) ` 

Theo ` PT, n_(Mg) = n_(H_2) = 0,15 ( mol ) `

` %m_(Mg) = ( 0,15 . 24 ) / ( 19,6 ) . 100 = 18,36 ( % ) `

` %m_(Fe_2O_3) = 100 - 18,36 = 81,64 ( % ) ` 

` 4,1 ) `

` m_(Mg) = 0,15 . 24 = 3,6 ( g ) `

` m_(Fe_2O_3) = 19,6 - 3,6 = 16 ( g ) `

` => n_(Fe_2O_3) = ( 16 ) / ( 160 ) = 0,1 ( mol ) `

` n_(HCl) = 0,15 . 2 + 0,1 . 6 = 0,9 ( mol ) `

` m_(HCl) = 0,9 . 36,5 = 32,85 ( g ) ` 

` mdd_(HCl) = ( 32,85 . 100 ) / ( 15 ) = 219 ( g ) ` 

` 4,2 ) ` 

` m_1 = m_(MgCl_2) + m_(FeCl_3) = 0,15 . 95 + ( 0,1 . 2 ) . 162,5 = 46,75 ( g ) `