`Zn + 2HCl -> ZnCl_2 + H_2`

`0,1`    `0,2`                           `0,1`     `(mol)`

`n_[Zn]=[6,5]/65=0,1(mol)`

`a)V_[H_2]=0,1.22,4=2,24(l)`

`b)C%_[HCl]=[0,2.36,5]/200 . 100 =3,65%`

`Zn + HCl -> ZnCl_2 + H_2` `\uparrow`

`n_(Zn) = (6,5)/65 = 0,1 mol`.

`n_(H_2) = 0,1 mol`.

`V(H_2) = 0,1 xx 22,4 = 2,24l`.

`C%(HCl) = (0,2.36,5)/200 xx 100 = 36,5%`.

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

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

0,1                     0,1         0,1 

\(b,C_M=\dfrac{0,1}{0,1}=1M\)

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

a)

$Fe + 2HCl \to FeCl_2 + H_2$
n H2 = n Fe = 11,2/56 = 0,2(mol)

V H2 = 0,2.22,4 = 4,48(lít)

b)

n HCl = 2n Fe = 0,2.2 = 0,4(mol)

=> CM HCl = 0,4/0,4 = 1M

c)

$CuO + H_2 \xrightarrow{t^o} Cu + H_2O$

Ta thấy :

n CuO = 64/80 = 0,8 > n H2 = 0,2 nên CuO dư

Theo PTHH :

n CuO pư = n Cu = n H2 = 0,2(mol)

n Cu dư = 0,8 - 0,2 = 0,6(mol)

Vậy : 

%m Cu = 0,2.64/(0,2.64 + 0,6.80)   .100% = 21,05%
%m CuO = 100% -21,05% = 78,95%

\(a,n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

PTHH: Zn + 2HCl ---> ZnCl₂ + H₂

          0,2-->0,4--------->0,2--->0,2

=> V = 0,2.22,4 = 4,48 (l)

b, Thiếu V dd

\(c,m_{ZnCl_2}=0,2.136=27,2\left(g\right)\)

a) 

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

            0,2-->0,4----->0,2--->0,2

=> V_H2 = 0,2.22,4 = 4,48 (l)

b) m_HCl = 0,4.36,5 = 14,6 (g)

=> \(m_{dd.HCl}=\dfrac{14,6.100}{7,3}=200\left(g\right)\)

c)

m_{dd sau pư} = 13 + 200 - 0,2.2 = 212,6 (g)

m_ZnCl2 = 0,2.136 = 27,2 (g)

=> \(C\%=\dfrac{27,2}{212,6}.100\%=12,8\%\)

\(n_{H_2}=\dfrac{4.48}{22.4}=0.2\left(mol\right)\)

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

\(...........0.4.....0.2...0.2\)

\(m_{HCl}=0.4\cdot36.5=14.6\left(g\right)\)

\(m_{MgCl_2}=0.2\cdot111=22.2\left(g\right)\)

\(^nFe=\dfrac{8,4}{56}=0,15\left(mol\right)\)

         \(Fe+2HCl\rightarrow FeCl_2+H_2\)

mol  0,15                   0,15      0,15

a)    \(V_X=V_{H_2}=0,15.22,4=3,36\left(l\right)\)

b)   \(^mFeCl_2=0,15.127=19,05\left(g\right)\)

Chúc bạn học tốt!!! 

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

a)\(n_{Fe}=0,15mol\Rightarrow n_{M_2}=0,15mol\Rightarrow V=0,15.22,4=3,36l\)

b)\(n_{FeCl_2}=n_{Fe}=0,15mol\Rightarrow m_{muối}=0,15.127=19,05g\)

\(a,Zn+2HCl\to ZnCl_2+H_2\\ b,n_{Zn}=\dfrac{13}{65}=0,2(mol);n_{HCl}=\dfrac{21,9}{36,5}=0,6(mol)\)

Vì \(\dfrac{n_{Zn}}{1}<\dfrac{n_{HCl}}{2}\) nên \(HCl\) dư

\(n_{HCl(dư)}=0,6-0,2.2=0,2(mol)\\ c,n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,2(mol)\\ \Rightarrow m_{ZnCl_2}=0,2.136=27,2(g)\\ V_{H_2}=0,2.22,4=4,48(l)\)