Question

Answer 1

a) nH2 = 0,195 (mol)

PTHH : \(Mg+2HCl-->MgCl_2+H_2\)

           \(2Al+6HCl-->2AlCl_3+3H_2\)

Đặt \(\left\{{}\begin{matrix}n_{Mg}=x\\n_{Al}=y\end{matrix}\right.\) => \(\left\{{}\begin{matrix}24x+27y=3,87\\x+\dfrac{3}{2}y=0,195\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=0,06\\y=0,09\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}m_{Mg}=1,44\left(g\right)\\m_{Al}=2,43\left(g\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\%m_{Mg}=37,2\%\\\%m_{Al}=62,8\%\end{matrix}\right.\)

b) nHCl pứ = 2nH2 = 0,39 (mol)

=> nHCl dư = 0,39.5/100 = 0,0195 (mol)

BT Mg : nMgCl2 = nMg = 0,06 (mol)

BT Al : nAlCl3 = nAl = 0,09 (mol)

=> \(C_{M\left(MgCl2\right)}=\dfrac{0,06}{0,5}=0,12\left(M\right)\)

   \(C_{M\left(AlCl_3\right)}=\dfrac{0,09}{0,5}=0,18\left(M\right)\)

   \(C_{M\left(HCl\right)}=\dfrac{0,0195}{0,5}=0,039\left(M\right)\)