Bài 4 :

`n_{Fe}=(5,6)/(56)=0,1(mol)`

`n_{HCl}=0,1.1=0,1(mol)`

`a,`PTHH:

`Fe+2HCl->FeCl_2+H_2`

Ta có tỉ lệ : `(0,1)/1 > (0,1)/2 `

`=>` Sau phản ứng `Fe` dư, `HCl` hết

`=>n_{H_2}= 1/2 n_{HCl}=1/2 .0,1=0,05(mol)`

`V_{H_2}=0,05.22,4=1,12(lít)`

`b,`

`n_{Fe (dư)}=0,1- (0,1)/2 =0,05(mol)`

`=>m_{Fe(dư)}=0,05.56=2,8(g)`

`c,`

`n_{FeCl_2}= 1/2 n_{HCl}= 1/2 . 0,1=0,05(mol)`

`=>C_{M FeCl_2}=(0,05)/(0,1)=0,5(M)`

Bài 5 :

`m_{NaOH}=200.20\%=40(g)`

`=>n_{NaOH}=(40)/(40)=1(mol)`

PTHH:

`NaOH+HCl->NaCl+H_2O`

    `1`  `->1`   `->1(mol)`

`a,``n_{NaCl}=1(mol)`

`=>m_{NaCl}=1.58,5=5,85(g)`

`C_{\% NaCl}=(5,85)/(200+100).100\%=1,95\%`

`b,`

`n_{HCl}=1(mol)`

`=>m_{HCl}=1.36,5=3,65(g)`

`C_{\% HCl}=(36,5)/(100).100\%=36,5\%`

Đáp án+ Giải thích các bước giải:

Bài 4:

Đổi `100(ml)=0,1(l)`

`n_{HCl}=C_MxxV=1xx0,1=0,1(mol)`

`n_{Fe}=m/M={5,6}/56=0,1(mol)`

`PT : Fe + 2HCl -> FeCl_2 + H_2`

`a)`

Tỉ lệ : `{0,1}/1>{0,1}/2-> Fe` dư

Theo pt : `n_{H_2}=1/2xx``n_{HCl}=1/2xx``0,1=0,05(mol)`

`->V_{H_2}=nxx22m4=0,05xx22,4=1,12(l)`

`b)`

Theo pt : `n_{Fe(pứ)}=n_{H_2}=0,05(mol)`

`->n_{Fe(dư)}=0,1-0,05=0,05(mol)`

`->m_{Fe(dư)}=nxxM=0,05xx56=2,8(g)`

`c)`

Theo pt : `n_{FeCl_2}=n_{H_2}=0,05(mol)`

`->` $C_{\text{M dd $FeCl_2$}}$ `=n/V={0,05}/{0,1}=0,5M`

Bài 5:

`n_{NaOH}={m_{dd}xxC%}/{100%xxM}={200xx20%}/{100%xx40}=1(mol)`

`PT : NaOH +  HCl -> NaCl + H_2O`

`a)`

Theo pt : `n_{NaCl}=n_{NaOH}=1(mol)`

`->C%`$_{\text{dd NaCl}}$ `={m_{ct}xx100%}/{m_{dd}}={1xx58,5xx100%}/{200+100}=19,5%`

`b)`

Theo pt : `n_{HCl}=n_{NaOH}=1(mol)`

`->C%`$_{\text{dd HCl}}$ `={m_{ct}xx100%}/{m_{dd}}={1xx36,5xx100%}/{100}=36,5%`

$\text{#Cyrus}$