a)

TN1:

FeO+H_2   $$\xrightarrow{t^o}$$  Fe+H_2O

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

Bảo toàn KL: m_A+m_{H_2}=m_B+m_{H_2O}

->18n_{H_2O}-2n_{H_2}=26,9-24,5=2,4(g)

Theo PT: n_{H_2O}=n_{H_2}={2,4}/{18-2}=0,15(mol)

->m_{H_2}=0,15.2=0,3(g)

TN2:

FeO+2HCl->FeCl_2+H_2O

CuO+2HCl->CuCl_2+H_2O

Al_2O_3+6HCl->2AlCl_3+3H_2O

b)

TN1:

Đặt n_{FeO}=x(mol);n_{CuO}=y(mol);n_{Al_2O_3}=z(mol)

->m_A=72x+80y+102z=26,9(1)

Theo PT: n_{Fe}=x(mol);n_{Cu}=y(mol)

->m_B=56x+64y+102z=24,5(2)

TN2:

Đặt n_{FeO}=kx(mol);n_{CuO}=ky(mol);n_{Al_2O_3}=kz(mol)

->kx+ky+kz=0,3(3)

Theo PT: n_{HCl}=2kx+2ky+6kz={300.14,6\%}/{36,5}=1,2(4)

(3)(4)->{x+y+z}/{2x+2y+6z}={0,3}/{1,2}=0,25

->x+y+z=0,5x+0,5y+1,5z

->x+y-z=0(5)

(1)(2)(5)->x=0,05;y=0,1;z=0,15

->\%m_{FeO}={0,05.72}/{26,9}.100\%\approx 13,38\%

        \%m_{CuO}={0,1.80}/{26,9}.100\%\approx 29,74\%

        \%m_{Al_2O_3}=100-29,74-13,38=56,88\%