Giả sử hỗn hợp X có khối lượng 100 (g)
Gọi số mol Fe, Mg là a, b (mol)
=> 56a + 24b = 100 (1)
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
a----->2a-------->a---->a
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
b------>2b------>b----->b
nHCl = 2a + 2b (mol)
=> mHCl = 73a + 73b (g)
=> \(m_{dd.HCl}=\dfrac{73a+73b}{15,91\%}=458,831a+458,831b\left(g\right)\)
mdd sau pư = 100 + 458,831(a+b) - 2(a+b)
= 100 + 456,831(a+b) (g)
Ta có: \(C\%_{FeCl_2}=\dfrac{127a}{100+456,831\left(a+b\right)}.100\%=15,239\%\)
=> \(\left\{{}\begin{matrix}a=1,389\left(mol\right)\\b=0,926\left(mol\right)\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{1,389.56}{100}.100\%=77,78\%\\\%m_{Mg}=\dfrac{0,926.24}{100}.100\%=22,22\%\end{matrix}\right.\)