a)

Đặt n_{Fe}=x(mol);n_{Zn}=y(mol)

->56x+65y=29,8(1)

Fe+2HCl->FeCl_2+H_2

Zn+2HCl->ZnCl_2+H_2

Theo PT: n_{H_2}=x+y={11,2}/{22,4}=0,5(2)

(1)(2)->x=0,3(mol);y=0,2(mol)

->\%m_{Fe}={0,3.56}/{29,8}.100\%\approx 56,38\%

->\%m_{Zn}=100-56,38=43,62\%

b)

Theo PT: n_{HCl}=2n_{H_2}=1(mol)

600ml=0,6l

->C_{M\ HCl}=1/{0,6}\approx 1,67M

c)

Cách 1:

Theo PT: n_{FeCl_2}=x=0,3(mol);n_{ZnCl_2}=y=0,2(mol)

->m_{m u o i}=0,3.127+0,2.136=65,3(g)

Cách 2:

BTKL: m_{m u o i}=29,8+1.36,5-0,5.2=65,3(g)