1.

a)

m_{Cu}=2(g)

n_{Al}=x(mol);n_{Fe}=y(mol)

->27x+56y=13,2-2=11,2(1)

2Al+6HCl->2AlCl_3+3H_2

Fe+2HCl->FeCl_2+H_2

->n_{H_2}=1,5x+y={5,6}/{22,4}=0,25(2)

(1)(2)->x=0,05(mol);y=0,175(mol)

->m_{Al}=0,05.27=1,35(g);m_{Fe}=11,2-1,35=9,85(g)

->\%m_{Al}=10,23\%;\%m_{Fe}=74,62\%;\%m_{Cu}=15,15\%

b)

2Al+3Cl_2  $$\xrightarrow{t^o}$$  2AlCl_3

2Fe+3Cl_2  $$\xrightarrow{t^o}$$  2FeCl_3

Cu+Cl_2  $$\xrightarrow{t^o}$$  CuCl_2

->n_{Cl_2}=1,5x+1,5y+2/{64}=0,36875(mol)

->V_{Cl_2}=0,36875.22,4=8,26(l)

2.

m_{Cu}=12,8(g)

n_{Mg}=x(mol);n_{Fe}=y(mol)

->24x+56y=23,6-12,8=10,8(1)

Mg+2HCl->MgCl_2+H_2

Fe+2HCl->FeCl_2+H_2

->n_{HCl}=2x+2y={91,25.20\%}/{36,5}=0,5(2)

(1)(2)->x=0,1;y=0,15

->\%m_{Mg}=10,17\%;\%m_{Cu}=54,24\%;\%m_{Fe}=35,59\%

n_{MgCl_2}=x=0,1(mol);n_{FeCl_2}=y=0,15(mol)

n_{H_2}=x+y=0,25(mol)

->V_{H_2}={0,25.0,082.(273+25)}/2=3,0545(l)

m_{dd\ spu}=10,8+91,25-0,25.2=101,55(g)

->C\%_{MgCl_2}={0,1.95}/{101,55}.100\%\approx 9,35\%

        C\%_{FeCl_2}={0,15.127}/{101,55}.100\%\approx 18,76\%