Giải hộ mình với ạ mmmmmm

Đáp án:

Giải thích các bước giải:

 a, 

`n_{H_2}=(11, 2) /(22, 4) =0, 5(mol) `

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

`a -> 3a -> a   -> 1,5a (mol) `

`Mg+2HCl->MgCl_2+H_2`

`b-> 2b-> b-> b(mol) `

$\rightarrow \begin{cases} 1,5a+b=0, 5\\27a+24b=10,2\end{cases}$

`=> a=b=0, 2`

`=> m_{Al}=0, 2.27=5, 4g`

`m_{Mg}=0, 2.24=4, 8g`

b, 

Theo pt:

`n_{HCl}=2n_{H_2}=1(mol) `

`=> n_{HCl dùng}=1+1.10%=1, 1(mol) `

`=>V_{dd HCl}=(1, 1) /1=1, 1L`

c,

Theo pt:

`n_{MgCl_2}=n_{Mg}=0, 2(mol) `

`n_{AlCl_3}=n_{Al}=0, 2(mol) `

`n_{HCl dư}=1, 1-1=0, 1(mol) `

`=> C_M MgCl_2=C_M AlCl_3=(0, 2)/(1, 1) ~~ 0,182M`

`C_M HCl=(0, 1) /(1, 1) ~~0, 091M`

`a)`

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

`Mg+2HCl->MgCl_2+H_2`

`Hh:Al:x(mol);Mg:y(mol)`

`->27x+24y=10,2(1)`

`n_{H_2}=1,5x+y={11,2}/{22,4}=0,5(2)`

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

`m_{Mg}=0,2.24=4,8(g)`

`m_{Al}=0,2.27=5,4(g)`

`b)`

`n_{HCl\ pu}=2n_{H_2}=1(mol)`

`n_{HCl\ du}=1.10\%=0,1(mol)`

`V_{dd\ HCl}={1+0,1}/{1}=1,1(L)`

`c)`

`n_{MgCl_2}=x=0,2(mol);n_{AlCl_3}=y=0,2(mol)`

`C_{M\ MgCl_2}=C_{M\ AlCl_3}={0,2}/{1,1}\approx 0,18M`

`C_{M\ HCl\ du}={0,1}/{0,18}=0,56M`