1. 

n_{HCl}=(52,14.1,05.10%)/(35,5)\approx 0,15(mol)

Bảo toàn H: n_{HCl}=2n_{H_2O}

->n_{H_2O}=(0,15)/(2)=0,075(mol)

->n_{O(Fe_xO_y)}=n_{H_2O}=0,075(mol)

->n_{Fe}=4-0,075.16=2,8(g)

x:y=n_{Fe}:n_O=(2,8)/(56):0,075=2:3

->Fe_2O_3

2.

a)

Bảo toàn H: n_{H_2SO_4}=n_{H_2O}=0,05.0,5=0,025(mol)

Bảo toàn O: n_{O(Fe_xO_y)}=n_{H_2O}=0,025(mol)

m>m_B=3,48(g)

->m_O+m_{NO_3^-}=3,48

->16.0,025+62n_{NO_3^-}=3,48

->n_{NO_3^-}=0,05(mol)

->n_{Fe(NO_3)_3}=(0,05)/(3)=(1)/(60)(mol)

Bảo toàn Fe: x.n_{Fe_xO_y}=n_{Fe(NO_3)_3}=(1)/(60)(mol)

->x:y=n_{Fe}:n_O=(1)/(60):0,025=2:3

b)

Áp dụng quy tắc đường chéo:

(n_{CO})/(n_{CO_2})=(44-17.2)/(17.2-28)=(5)/(3)

->%CO=(5)/(5+3).100%=62,5%

->%CO_2=100%-62,5%=37,5%

c)

m=(1)/(60).56+0,025.16=(4)/(3)(g)

n_{CO(P,ứ)}=n_{CO_2}=n_{O(Fe_2O_3}}=0,025(mol)

->n_{CO(dư)}=x-0,025(mol)

M=(0,025.44+(x-0,025).28)/(x)=17.2

->x=(1)/(15)

->V=(1)/(15).22,4\approx 1,5(l)

a)

n{HCl}={52,14.1,05.10\%}/{36,5}=0,15(mol)

Fe_xO_y+2yHCl->xFeCl_{2y//x}+yH_2O

->n_{Fe_xO_y}={0,15}/{2y}={0,075}/{y}(mol)

->M_{Fe_xO_y}=4/{{0,075}/y}={160}/{3}y

->56x+16y={160}/{3}y

->x/y=2/3

->Fe_xO_y:\ Fe_2O_3

b)

Fe_2O_3+3CO  $$\xrightarrow{t^o}$$  2Fe+3CO_2

->{n_{CO\ du}}/{n_{CO_2}}={44-17.2}/{17.2-28}=5/3

->\%V_{CO\ du}=5/{5+3}.100\%=62,5\%

->\%V_{CO_2}=100-62,5=37,5\%

c)

n_{H_2SO_4}=0,05.0,5=0,025(mol)

Quy B về Fe:x(mol);Fe_2O_3\ du:y(mol)

->m_B=56x+160y(g)

Fe+H_2SO_4->FeSO_4+H_2

Fe_2O_3+3H_2SO_4->Fe_2(SO_4)_3+3H_2O

Theo PT: n_{H_2SO_4}=x+3y=0,025(1)

Fe+4HNO_3->Fe(NO_3)_3+NO+2H_2O

Fe_2O_3+6HNO_3->2Fe(NO_3)_3+3H_2O

Theo PT: n_{Fe(NO_3)_3}=x+2y(mol)

->m_{Fe(NO_3)_3}=242(x+2y)=56x+160y+3,48(2)

(1)(2)->x=0,01;y=0,005

Fe_2O_3+3CO  $$\xrightarrow{t^o}$$  2Fe+3CO_2

->n_{CO\ pu}=n_{CO_2}=3/2x=0,015(mol)

->n_{CO\ du}=0,015.5/3=0,025(mol)

->V=(0,015+0,025).22,4=0,896(l)

Theo PT: n_{Fe_2O_3\ pu}=1/2x=0,005(mol)

->m=(0,005+0,005).160=1,6(g)