Đáp án+ Giải thích các bước giải:

Bài 3:

`V_{dd(sau)}=2,5+3=5,5(l)`

`n_{dd(sau)}=2,5xx0,5+3xx1=4,25(mol)`

`->` $C_{\text{M dd(sau)}}$`=n/V={4,25}/{5,5}=0,772M`

Bài 4:

`a)`

`PT : Zn + 2HCl -> ZnCl_2 + H_2`

`b)`

`n_{HCl}={m_{dd}xxC%}/{100%xxM}={50xx7,3%}/{100%xx36,5}=0,1(mol)`

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

`->V_{H_2}=nxx22,4=0,05xx22,4=1,12(l)`

`c)`

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

`->m_{Zn}=nxxM=0,05xx65=3,25(g)=m`

Bài 5:

Đổi `50(ml)=0,05(l)`

`n_{H_2SO_4}=C_MxxV=2xx0,05=0,1(mol)`

`->m_{H_2SO_4}=nxxM=0,1xx98=9,8(g)`

$\text{#Cyrus}$

$3)$

`n_(đg)(1)=2,5.0,5=1,25(mol)`

`n_(đg)(2)=3.2=3(mol)`

`∑n_(đg)=1,25+3=4,25(mol)`

`∑V_(đg)=2,5+3=5,5(l)`

`=>CM_(đg)=(4,25)/(5,5)=0,77(M)`

$4)$

`m_(ctHCl)=50.7,3%=3,65(g)`

`=>n_(HCl)=(3,65)/(36,5)=0,1(mol)`

$a)PTHH :$

$Zn+2HCl \to ZnCl_2+H_2$

$b)$

Theo `PT,n_(H_2)=1/2n_(HCl)=0,05(mol)`

`=>V_(H_2)(đktc)=0,05.22,4=1,12(l)`

$c)$

Theo `PT,n_(Zn)=1/2n_(HCl)=0,05(mol)`

`m=m_(Zn)=0,05.65=3,25(g)`

$5)$

Đổi $50(ml)=0,05(l)$

`=>n_(H_2SO_4)=0,05.2=0,1(mol)`

`=>m_(H_2SO_4)=0,1.98=9,8(g)`