Question

Answer 1

Gọi \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_M=b\left(mol\right)\end{matrix}\right.\)

=> 56a + b.M_M = 11,3 (1)

n_HCl = 0,3.2,5 = 0,75 (mol)

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

=> Axit còn dư

PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

             a------------------------>a

            \(M+2HCl\rightarrow MCl_2+H_2\)

              b----------------------->b

=> a + b = 0,3 

(1) => 56(0,3-b) + b.M_M = 11,3

=> b(56 - M_M) = 5,5 

Mà b < 0,3

=> M_M < 37,67 (*)

\(n_{H_2SO_4}=0,2.2=0,4\left(mol\right)\) 

PTHH: \(M+H_2SO_4\rightarrow MSO_4+H_2\)

           0,4<-----0,4

=> \(\dfrac{4,8}{M_M}< 0,4\Rightarrow M_M>12\) (**)

(*)(**) => M là Mg

Ta có: \(\left\{{}\begin{matrix}56a+24b=11,3\\a+b=0,3\end{matrix}\right.\)

=> a = 0,128125 (mol); b = 0,171875 (mol)

=> \(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,128125.56}{11,3}.100\%=63,5\%\\\%m_{Mg}=\dfrac{0,171875.24}{11,3}.100\%=36,5\%\end{matrix}\right.\)