nhh khi = 0,3 mol
MnO2 + 4HCl\(\rightarrow\) MnCl2 + Cl2 + H2O
a _____________________a mol
Mg + 2HCl \(\rightarrow\) MgCl2 + H2
b___________________b mol
Ta có : \(\left\{{}\begin{matrix}87a+24b=13,5\\a+b=0,3\end{matrix}\right.\rightarrow\left\{{}\begin{matrix}a=0,1\\b=0,2\end{matrix}\right.\)
\(\rightarrow\) % mMnO2 = \(\frac{\text{0,1.87}}{13,5}.100\%\) = 64,44%
\(\rightarrow\)% mMg = 100% - 64,44% = 35,56%