Đặt `n_(Al) = x; n_(Cu) = y; n_(Fe) = z`

Thí nghiệm 1

`m_A = 27x + 64y + 56z = 31,1g(1)`

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

`Fe + 2HCl -> FeCl_2 + H_2`

Theo PTHH

`n_(H_2) = 3/2n_(Al) + n_(Fe)`

`=>(0,784)/(22,4) = 0,035 = 3/2x + z(2)`

Thí nghiệm 2

Đặt `n_(Al) = kx; n_(Cu) = ky; n_(Fe) = kz`

`=> k(x+y+z) = 0,12(3)`

`2Al + 3Cl_2 \overset(t^@)(->) 2AlCl_3`

`Cu + Cl_2  \overset(t^@)(->) CuCl_2`

`2Fe + 3Cl_2  \overset(t^@)(->) 2FeCl_3`

Theo PTHH:

`n_(AlCl_3) = n_(Al) = kx(mol)`

`n_(CuCl_2) = n_(Cu) = ky(mol)`

`n_(FeCl_3) = n_(Fe) = kz(mol)`

`=> 133,5kx + 135ky + 162,5kz  = 17,27`

`=> k(133,5x + 135y + 162,5z) = 17,27(4)`

Lấy `(3)` chia `(4)`

`=> (k(x + y + z))/(k(133,5x + 135y + 162,5z)) = (0,12)/(17,27)`

`=> (x + y + z)/(133,5x + 135y + 162,5z) = (12)/(1727)`

`=> 1727x + 1727y + 1727z = 1602x + 1620y + 1950z`

`=> 125x +  107y - 223z = 0(5)`

`(1),(2),(5) => x ~~ -0,073;y~~ 0,39;z ~~ 0,146`

`=>` Xem lại đề.