\(\left\{{}\begin{matrix}Al:x\left(mol\right)\\Fe:y\left(mol\right)\end{matrix}\right.\)→ 27x + 56y = 22(1)
\(2Al + 3Cl_2 \xrightarrow{t^o} 2AlCl_3\\ 2Fe + 3Cl_2 \xrightarrow{t^o} 2FeCl_3\)
\(\left\{{}\begin{matrix}AlCl_3:x\left(mol\right)\\FeCl_3:y\left(mol\right)\end{matrix}\right.\)→ 133,5x + 162,5y = m1
\(2Al +3 I_2 \xrightarrow{t^o} 2AlI_3\\ Fe + I_2 \xrightarrow{t^o} FeI_2\)
\(\left\{{}\begin{matrix}AlI_3:x\left(mol\right)\\FeI_2:y\left(mol\right)\end{matrix}\right.\) → 408x + 310y = m2
Suy ra : m2 - m1 = 274,5x + 147,5y = 139,3(2)
Từ (1)(2) suy ra: x = 0,4 ; y = 0,2
Vậy :
mAl = 0,4.27 = 10,8(gam)
mFe = 0,2.56 = 11,2(gam)