Gọi a, b, c là mol Fe, Cu, Mg trong 20,16g hh.
$\Rightarrow 56a+64g+24c=20,16$ (1)
$n_{H_2}= 0,36 mol$
Bảo toàn e: $2a+2c=0,36.2=0,72$ (2)
Trong 0,4 mol hh có ka, kb, kc mol Fe, Cu, Mg.
$\Rightarrow k(a+b+c)=0,4$ (*)
$n_{Cl_2}=0,45 mol$
Bảo toàn e: $k(3a+2b+2c)=0,45.2=0,9$ (**)
(*)(**) $\Rightarrow k=\frac{0,4}{a+b+c}=\frac{0,9}{3a+2b+2c}$
$\Leftrightarrow 0,9a+0,9b+0,9c=0,4(3a+2b+2c)$
$\Leftrightarrow -0,3a+0,1b+0,1c=0$ (3)
(1)(2)(3) $\Rightarrow a=b=0,12; c=0,24$
$m_{Fe}= 0,12.56=6,72g$
$m_{Cu}=0,12.64=7,68g$
$m_{Mg}=0,24.24=5,76g$