a)

n_{HCl}=0,5.0,4=0,2(mol)

Đặt n_{Fe}=x(mol);n_{Zn}=y(mol)

Fe+2HCl->FeCl_2+H_2

Zn+2HCl->ZnCl_2+H_2

Theo PT:

n_{HCl}=2x+2y=0,2(1)

n_{FeCl_2}=x(mol);n_{ZnCl_2}=y(mol)

->127x+136y=13,42(2)

(1)(2)->x=0,02;y=0,08

->\%m_{Fe}={0,02.56}/{0,02.56+0,08.65}.100\%\approx 17,72\%

->\%m_{Zn}=100-17,72=82,28\%

b) A chỉ chứa muối trung hòa, không thể dùng bazo để trung hòa tiếp được.

{(n_(Fe)=x(mol)),(n_(Zn)=y(mol)):}

n_(HCl)=0,5.0,4=0,2(mol)

Fe+2HCl->FeCl_2+H_2

Zn+2HCl->ZnCl_2+H_2

=>2x+2y=0,2

{(n_(FeCl_2)=x(mol)),(n_(ZnCl_2)=y(mol)):}

=>127x+136y=13,42

=>{(x=0,02),(y=0,08):}

=>m_(Fe)=0,02.56=1,12(g)

m_(Zn)=0,08.65=5,2(g)

=>%m_(Fe)=(1,12)/(1,12+5,2).100%=17,72%

=>%m_(Zn)=100%-17,72%=82,28%

b

-> sai đề