1)

a)`2Al + 6HCl -> 2AlCl_3 + 3H_2(1)`

`Fe + 2HCl -> FeCl_2 + H_2(2)`

b)`n_{H_2} = {8,96}/{22,4} = 0,4` mol

Theo PTHH (1), (2)

`n_{HCl} = 2n_{H_2} = 2 . 0,4 = 0,8` mol

`m = m_{dd HCl} = {0,8 . 36,5 . 100}/{14,6} = 200g`

c)đặt `n_{Al} = x(mol), n_{Fe} = y(mol)`

`=> 27x + 56y = 11(I)`

Theo PT(1), (2)

`n_{H_2} = 1,5n_{Al} + n_{Fe} = 1,5x + y = 0,4(II)`

`(I), (II) => x = 0,2; y = 0,1`

`%m_{Al} = {0,2 . 27}/{11} . 100% ~~ 49,1%`

`%m_{Fe} = 100% - 49,1% = 50,9%`

C2:

Gọi `n` là hóa trị của kim loại `R`

`2R + nCl_2 ->^{t^o} 2RCl_n`

a)Bảo toàn khối lượng

`=> m_{Cl_2} = 26,7 - 5,4 = 21,3g`

`n_{Cl_2} = {21,3}/{71} = 0,3` mol

`V = V_{Cl_2} = 0,3. 22,4 = 6,72l`

b)`n_R = 2/n n_{Cl_2} = {0,3 . 2}/{n} = {0,6}/{n}` mol

`M_R = {5,4}/{{0,6}/{n}} = 9n(g//mol)`

`n = 3 => M_R = 27(Al)`

Vậy R là kim loại `Al`

`1)`

`a)`

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

`Fe+2HCl->FeCl_2+H_2`

`b)`

`X:Al:x(mol);Fe:y(mol)`

`27x+56y=11(1)`

`n_{H_2}={8,96}/{22,4}=0,4(mol)`

Theo pt:

`n_{HCl}=2n_{H_2}=0,8(mol)`

`m={0,8.36,5}/{14,6\%}=200(g)`

`c)`

Theo pt:

`n_{H_2}=1,5x+y=0,4(2)`

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

`\%m_{Al}={0,2.27}/{11}.100\%=49,09\%`

`\%m_{Fe}=100-49,09=50,91\%`

`2)`

`a)`

`X` hóa trị n

`2X+nCl_2` $\xrightarrow{t^o}$ `2XCl_n`

BTKL: `n_{Cl_2}={26,7-5,4}/{71}=0,3(mol)`

`V=0,3.22,4=6,72(L)`

`b)`

Theo pt:

`n_X={2n_{Cl_2}}/n={0,6}/n(mol)`

`M_X={5,4}/{{0,6}/n}=9n`

`->n=3;M_X=27(Al)`