Zn + 2HCl -> ZnCl_2 + H_2(1)

n_{Zn} = {m}/{65}(mol)

a)Theo PTHH(1)

n_{H_2} = n_{Zn} = {m}/{65}(mol)

Ta có:

m_{tăng} = m_{Zn} - m_{H_2}

25,2 = m - {2m}/{65}

=> m = 26

b)n_{H_2} = m/{65} = 0,4 mol

CuO + H_2 ->^{t^o} Cu + H_2O(2)

Fe_3O_4 + 4H_2 ->^{t^o} 3Fe + 4H_2O(3)

Đặt n_{CuO} = x(mol)

=> n_{Cu} = n_{CuO} = x(mol)

n_{Fe_3O_4} = y(mol)

=> n_{Fe} = 3n_{Fe_3O_4} = 3y(mol)

Theo PT(2),(3)

n_{H_2} = n_{CuO} + 4n_{Fe_3O_4} = x + 4y = 0,4 mol(I)

m_{hh rắn} = m_{Cu} + m_{Fe} = 64x + 56 . 3y = 19g(II)

(I), (II) => x = 0,1; y = 0,075

%m_{Cu} = {0,1 . 64}/{19} . 100% $$\approx$$ 33,684%

%m_{Fe} = 100% - 33,684% = 66,316%

a)

Zn+2HCl->ZnCl_2+H_2

->m_{dd\ tang}=m_{Zn}-m_{H_2}=25,2(g)

->65n_{Zn}-2n_{H_2}=25,2

Mà theo PT: n_{Zn}=n_{H_2}

->n_{Zn}=n_{H_2}=0,4(mol)

->m_{Zn}=0,4.65=26(g)

b)

H_2+CuO  $$\xrightarrow{t^o}$$  Cu+H_2O

4H_2+Fe_3O_4  $$\xrightarrow{t^o}$$  3Fe+4H_2O

Đặt n_{Cu}=x(mol);n_{Fe}=y(mol)

->64x+56y=19(1)

Theo PT: n_{H_2}=x+4/{3}y=0,4(2)

(1)(2)->x=0,1;y=0,225

\%m_{Cu}={0,1.64}/{19}.100\%\approx 33,68\%

\%m_{Fe}=100-33,68=66,32\%