a)

gọi a là số mol Mg b là số mol Al

Ta có

24a+27b=7,65

a+1,5b=0,375
giải hệ pt 

=>a=0,15 b=0,15

%mMg=0,15x24/7,65x100%=47,06%

%mAl=100-47,06=52,94%

b)

nHCl=0,375x2=0,75 mol

mdd HCl=0,75x36,5/10%=273,75 g

c)

mdd spu=7,65+273,75-0,375x2=280,65 g

C%MgCl2=0,15x95/280,65x100%=5,08%

C%AlCl3=0,15x133,5/280,65x100%=7,14%

Mg+2HCl→MgCl2+H2Mg+2HCl→MgCl2+H2

2Al+6HCl→2AlCl3+3H22Al+6HCl→2AlCl3+3H2

nH2=8,422,4=0,375(mol)nH2=8,422,4=0,375(mol)

a)

Gọi a là số mol Mg b là số mol Al

Ta có

{24a+27b=7,65a+1,5b=0,375→{a=0,15b=0,15{24a+27b=7,65a+1,5b=0,375→{a=0,15b=0,15

%mMg=0,15.247,65.100%=47,06%%mMg=0,15.247,65.100%=47,06%

%mAl=100%−47,06%=52,94%%mAl=100%−47,06%=52,94%

b)

nHCl=0,375.2=0,75(mol)nHCl=0,375.2=0,75(mol)

mddHCl=0,75.36,510%=273,75(g)mddHCl=0,75.36,510%=273,75(g)

c)

mddspu=7,65+273,75−0,375.2=280,65(g)mddspu=7,65+273,75−0,375.2=280,65(g)

C%MgCl2=0,15.95280,65.100%=5,08%C%MgCl2=0,15.95280,65.100%=5,08%

C%AlCl3=0,15.133,5280,65.100%=7,14%