8Al+3Fe_3O_4  $$\xrightarrow{t^o}$$  4Al_2O_3+9Fe

P1:

Đặt n_{Al}=x(mol);n_{Al_2O_3}=y(mol)

->n_{Fe}=2,25y(mol)

2Al+2NaOH+2H_2O->2NaAlO_2+3H_2

n_{H_2}={672.10^{-3}}/{22,4}=0,03(mol)

Theo PT: x=n_{Al}=2/3n_{H_2}=0,02(mol)

P2:

Đặt n_{Al}=0,02k(mol);n_{Al_2O_3}=ky(mol)

->n_{Fe}=2,25ky(mol)

Fe+2HCl->FeCl_2+H_2

2Al+6HCl->2AlCl_3+3H_2

Theo PT: 1,5n_{Al}+n_{Fe}={18,816}/{22,4}=0,84(mol)

->0,03k+2,25ky=0,84

->k={0,84}/{0,03+2,25y}

Xét toàn phần:

Theo PT:

n_{Al\ pu}=2n_{Al_2O_3}=2(ky+y)(mol)

n_{Fe_3O_4}=1/3n_{Fe}=1/3(2,25y+2,25ky)(mol)

->m_X=54ky+54y+174y+174ky+27(0,02k+0,02)=93,9

->228ky+228y+0,54k=93,36

->{191,52y}/{0,03+2,25y}+228y+{0,4356}/{0,03+2,25y}=93,36

->y\approx 0,08(mol)

->k=4

Vậy hh ban dau:

m_{Fe_3O_4}=1/{3}(2,25.0,08+2,25.4.0,08).232=69,6(g)

m_{Al}=93,9-69,6=24,3(g)