- ta có :

7,3% = \(\dfrac{m_{HCl}}{50}\).100%

\(\Rightarrow\) m_HCl = 3,65 g

\(\Rightarrow\)n_HCl = \(\dfrac{3,65}{36,5}\) = 0,1 mol

a) Zn + 2HCl \(\rightarrow\)ZnCl₂ + H₂ \(\uparrow\)

0,05..\(\leftarrow\)0,1................\(\rightarrow\)0,05

b) m = 0,05 . 65 = 3,25 g

V_H2 = 0,05 . 22,4 = 1,12 (l)

\(m_{HCl}=50.7,3\%=3,65\left(g\right)\\ n_{HCl}=\dfrac{3,65}{36,5}=0,1\left(mol\right)\\ PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{Zn}=n_{H_2}=n_{ZnCl_2}=\dfrac{1}{2}n_{HCl}=\dfrac{1}{2}.0,1=0,05\left(mol\right)\\ m_{Zn}=0,05.65=3,25\left(g\right)\\ V_{H_2\left(ĐKTC\right)}=0,05.22,4=1,12\left(l\right)\\ m_{ZnCl_2}=0,05.136=6,8\left(g\right)\)

mHCl=50.7,3%=3,65(g) -> nHCl=0,1(mol)

a) PTHH: Zn + 2 HCl -> ZnCl2 + H2

nH2=nZnCl2=nZn=nHCl/2= 0,1/2=0,05(mol)

b) m=mZn=0,05.65=3,25(g)

c) V(H2,đktc)=0,05.22,4=1,12(l)

d) mZnCl2= 136.0,05= 7,8(g)

\(n_{Zn}=\dfrac{1,3}{65}=0,02\left(mol\right)\\ pthhZn+2HCl\rightarrow ZnCl_2+H_2\)
        0,02                   0,02        0,02 
\(m_{ZnCl_2}=136.0,02=2,72\left(g\right)\\ V_{H_2}=0,02.22,4=0,448\left(l\right)\)

\(n_{Zn}=\dfrac{32,25}{65}=0,49mol\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

0,49                                 0,49  ( mol )

\(V_{H_2}=0,49.22,4=10,976l\)

a) Ta có PTHH sau: Zn + 2HCl ---> ZnCl2 + H2
b) Ta có: nZn = 32,25/65 ∼0,5(mol) 
=> nH2 = 0,5(mol)
=> V của H2 là: 0,5x22,4 = 11,2(l)
Chúc bn học tốt :)

`a) PTHH:`

`Zn + 2 HCl -> ZnCl_2 + H_2`

`0,05`    `0,1`                        `0,05`    `(mol)`

`n_[HCl] = [ [ 7,3 ] / 100 . 50 ] / [ 36,5 ] = 0,1 (mol)`

`b) V_[H_2] = 0,05 . 22,4 = 1,12 (l)`

`c) m_[Zn] = 0,05 . 65 = 3,25 (g)`

\(m_{HCl}=\dfrac{50.7,3}{100}=3,65g\\ n_{HCL}=\dfrac{3,65}{36,5}=0,1\left(mol\right)\\ pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\) 
          0,05    0,1                       0,05 
\(V_{H_2}=0,5.22,4=1,12l\\ m_{Zn}=0,05.65=3,25g\)

a) 

Zn  +  2HCl  → ZnCl₂  +  H₂

nHCl = 50.7,3% = 3,65 gam

<=> nHCl = 3,65 : 36,5 = 0,1 mol

Theo tỉ lệ phản ứng => nZn = 1/2nHCl = 0,05 mol 

<=> mZn = 0,05.65 = 3,25 gam.

b) nH₂ = 1/2nHCl = 0,05 mol 

=> VH₂ = 0,05 . 22,4 = 1,12 lít.

c) m dung dịch sau phản ứng = mZn + m dd HCl - mH₂ = 3,25 + 50 - 0,05.2 = 53,15 gam

mZnCl₂ = 0,05.136 = 6,8 gam

<=> C% _ZnCl2 = \(\dfrac{6,8}{53,15}.100\%\) = 12,8%

\(n_{HCl}=\dfrac{100.7,3\%}{36,5}=0,2\left(mol\right)\\ a,PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ b,n_{Zn}=n_{H_2}=n_{ZnCl_2}=\dfrac{0,2}{2}=0,1\left(mol\right)\\ m=m_{Zn}=0,1.65=6,5\left(g\right)\\ c,V_{H_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\\ d,m_{ddZnCl_2}=6,5+100-0,1.2=106,3\left(g\right)\\ C\%_{ddZnCl_2}=\dfrac{0,1.136}{106,3}.100\approx12,794\%\)

a. \(n_{Zn}=\dfrac{6.5}{65}=0,1\left(mol\right)\)

PTHH : Zn + 2HCl -> ZnCl₂ + H₂

            0,1      0,2                  0,1

b. \(V_{H_2}=0,1.22,4=2,24\left(l\right)\)

c. \(m_{HCl}=0,2.36,5=7,3\left(g\right)\)

\(n_{Zn}=\dfrac{6,5}{65}=0,1mol\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

0,1      0,2                       0,1

\(V_{H_2}=0,1\cdot22,4=2,24l\)

\(m_{HCl}=0,2\cdot36,5=7,3g\)