$$\begin{gathered} {\mathrm{nH}}_2=0,14 \mathrm{mol} \\ \mathrm{Mg}+2\mathrm{HCl}\to {\mathrm{MgCl}}_2+{\mathrm{H}}_2 \\ \mathrm{x} \mathrm{x} \left(\mathrm{mol}\right) \\ \mathrm{Fe}+2\mathrm{HCl}\to {\mathrm{FeCl}}_2+{\mathrm{H}}_2 \\ \mathrm{y} \mathrm{y} \left(\mathrm{mol}\right) \\ \mathrm{Ta} \mathrm{có}: \left\{\begin{array}{l}24\mathrm{x}+56\mathrm{y}=6,56\\ \mathrm{x}+\mathrm{y}=0,14\end{array}<=>\left\{\begin{array}{l}\mathrm{x}=0,04\\ \mathrm{y}=0,1\end{array}\right.\right. \\ \mathrm{Nếu} 6,56\mathrm{g} \mathrm{Fe}+\mathrm{Mg} \mathrm{tác} \mathrm{dụng} \mathrm{với} {\mathrm{Cl}}_2 \end{gathered}$$

$$\begin{gathered} \mathrm{Mg}+{\mathrm{Cl}}_2\stackrel{\mathrm{to}}{\to }{\mathrm{MgCl}}_2 \\ 0,04 0,04 0,04 \left(\mathrm{mol}\right) \\ 2\mathrm{Fe}+3{\mathrm{Cl}}_2\stackrel{\mathrm{to}}{\to }2{\mathrm{FeCl}}_3 \\ 0,1 0,1 \left(\mathrm{mol}\right) \\ =>{\mathrm{m}}_{\mathrm{muối}}={\mathrm{m}}_{{\mathrm{MgCl}}_2}+{\mathrm{m}}_{{\mathrm{FeCl}}_3}=20,05\mathrm{g} \\ \mathrm{Mà} \mathrm{theo} \mathrm{bài} \mathrm{tạo} \mathrm{ra} 50,125\mathrm{g} \mathrm{gấp} 2,5 \mathrm{lần} \\ =>\mathrm{m}=2,5.6,56=16,4\mathrm{g} \end{gathered}$$

Vậy m=16,4g

nH2=0,14 molMg+2HCl→MgCl2+H2  x                                   x      (mol)Fe+2HCl→FeCl2+H2  y                                  y        (mol)Ta có: {24x+56y=6,56x+y=0,14<=>{x=0,04y=0,1Nếu 6,56g Fe+Mg tác dụng với Cl2nH2=0,14 molMg+2HCl→MgCl2+H2  x                                   x      (mol)Fe+2HCl→FeCl2+H2  y                                  y        (mol)Ta có: 24x+56y=6,56x+y=0,14<=>x=0,04y=0,1Nếu 6,56g Fe+Mg tác dụng với Cl2

Mg+Cl2to→MgCl20,04 0,04       0,04     (mol)2Fe+3Cl2to→2FeCl30,1                      0,1        (mol)=>mmuối=mMgCl2+mFeCl3=20,05gMà theo bài tạo ra 50,125g gấp 2,5 lần=>m=2,5.6,56=16,4gMg+Cl2→toMgCl20,04 0,04       0,04     (mol)2Fe+3Cl2→to2FeCl30,1                      0,1        (mol)=>mmuối=mMgCl2+mFeCl3=20,05gMà theo bài tạo ra 50,125g gấp 2,5 lần=>m=2,5.6,56=16,4g

Vậy m=16,4g