Chỉ tui bài 82 với anh em giúp với

Giải thích các bước giải:

`a)`

`500ml=0,5l`

`=>n_(HCl)=0,5.2=1(mol)`

Đặt `n_(Mg)=a;n_(Al)=b(mol)`

`Mg+2HCl->MgCl_2+H_2`

`a->2a->a->a(mol)`

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

`b->3b->b->3/2b(mol)`

Có `n_(H_2) = (11, 1555)/(24,79)=0,45(mol)`

`BTNT(H):n_(H)=n_(HCl)=0,45.2=0,9<1(mol)`

`=>HCl` dư

`=>2a+3b=n_(HCl\ pứ)=0,9(mol)`

Có `m_(M)=40,95`

`=>95a+133,5b=40,95(2)`

`(1);(2)`

`=>a=0,15;b=0,2(mol)`

`=>x=0,15.24+0,2.27=9(g)`

`b)`

`C_(M\ MgCl_2) = (0,15)/(0,5)=0,3(M)`

`C_(M\ AlCl_3)=(0,2)/(0,5)=0,4(M)`

`C_(M\ HCl\ dư) = (1-0,9)/(0,5)=0,2(M)`

$\rm n_{HCl}=0,5.2=1(mol)$

$\rm n_{H_2}=\dfrac{11,1555}{24,79}=0,45(mol)$

$\rm PTHH:Mg+2HCl \to MgCl_2+H_2$

$\rm 2Al+6HCl \to 2AlCl_3+3H_2$

$\rm n_{HCl}=2n_{H_2}=0,9(mol)(<n_{HCl bđ})$

`=>`$\rm n_{Cl}=0,9(mol)$

$\rm \to HCl dư$

ta có : $\rm m_{muối}=x+m_{Cl}$

$\rm \to x=m_{muối}-m_{Cl}=40,95-0,9.35,5=9(g)$

$\rm b.$ gọi x,y là số mol $\rm Mg,Al$

$\rm \to 24x+27y=9(1)$
$\rm n_{H_2}=0,45(mol) \to x+1,5y=0,45(2)$

$\rm (1)(2) \to x=0,15;y=0,2$
$\rm C_M MgCl_2=\dfrac{0,15}{0,5}=0,3(M)$
$\rm C_M AlCl_3=\dfrac{0,2}{0,5}=0,4(M)$

$\rm C_M HCl dư=\dfrac{0,1}{0,5}=0,2(M)$