`Zn + 2HCl -> ZnCl_2 + H_2 \uparrow`

`0,25`  `0,5`                         `0,25`           `(mol)`

`n_[Zn]=[16,5]/65=0,25(mol)`

`a)m_[HCl]=0,5.36,5=18,25(g)`

`b)V_[H_2]=0,25.22,4=5,6(l)`

a)nZn= 6,5/65= 0,1 (mol)

Zn +2HCl --> ZnCl2 + H2 (1)

0,1------------0,1-------0,1

Theo (1), nH2= nZn= 0,1 (mol)

=> VH2= 0,1 . 22,4 = 2,24 (lít)

=>mZnCl2=0,1.136=13,6g

a)

$Zn + 2HCl \to ZnCl_2 + H_2$

b)

Theo PTHH : 

$n_{Zn} = n_{H_2} = \dfrac{5,6}{22,4} = 0,25(mol)$
$m_{Zn} = 0,25.65 =16,25(gam)$

c)

$n_{HCl} = 2n_{H_2} = 0,5(mol)$

Số phân tử $HCl = 0,5.6.10^{23} = 3.10^{23}$ phân tử

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

b. Theo ĐLBTKL, ta có:

\(m_{Zn}+m_{HCl}=m_{ZnCl_2}+m_{H_2}\\ b.\Leftrightarrow6,5+7,3=13,6+m_{H_2}\\ \Leftrightarrow m_{H_2}=\left(6,5+7,3\right)-13,6=0,2\left(g\right)\)

Chúc em học tốt!

\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ pthh:Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)  
          0,2       0,4           0,2 
\(m_{MgCl_2}=0,2.95=19\left(g\right)\\ C\%_{HCl}=\dfrac{0,4.36,5}{200}.100\%=7,3\%\)

nFe = 7,2 : 56 = 9/70 (mol ) 
pthh : Fe + 2HCl ---> FeCl2 + H2
         9/70-->9/35-------------->9/70 (mol) 
=> VH2 = 9/70. 22,4 =2,88(l)
=> mHCl = 9/35 . 36,5 = 9,38 (g) 

Số mol của sắt là 7,2/56=9/70 (mol).

a/ PTHH: Fe (9/70 mol) + 2HCl (9/35 mol) \(\rightarrow\) FeCl₂ + H₂\(\uparrow\) (9/70 mol).

b/ Thể tích khí hiđro sinh ra:

V=9/70.22,4=2,88 (lít).

c/ Khối lượng của axit HCl đã dùng:

m=9/35.36,5=657/70 (g)\(\approx\)9,386 (g).

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

PTHH: Zn + 2HCl → ZnCl₂ + H₂

Mol:    0,1                  0,1           0,1

b,\(m_{ZnCl_2}=0,1.136=13,6\left(g\right)\)

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

a)PTHH:  \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

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

\(n_{ZnCl_2}=n_{Zn}=0,1\left(mol\right)\)

\(\Rightarrow m_{ZnCl_2}=n.M=0,1.136=13,6\left(g\right)\)

c) \(n_{H_2}=n_{Zn}=0,1\left(mol\right)\)

\(V_{H_2}=n.22,4=0,1.22,4=2,24\left(l\right)\)

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

a) Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2|\)

            1         2              1          1

          0,1                        0,1      0,1

b) \(n_{ZnCl2}=\dfrac{0,1.}{1}=0,1\left(mol\right)\)

⇒ \(m_{ZnCl2}=01,.136=13,6\left(g\right)\)

c) \(n_{H2}=\dfrac{0,1.}{1}=0,1\left(mol\right)\)

\(V_{H2\left(dktc\right)}=01,.22,4=2,24\left(l\right)\)

 Chúc bạn học tốt

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right);n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\a, Zn+2HCl\rightarrow ZnCl_2+H_2\\ b,V\text{ì}:\dfrac{0,2}{1}>\dfrac{0,1}{1}\Rightarrow Zn\text{dư}\\ \Rightarrow n_{Zn\left(p.\text{ứ}\right)}=n_{ZnCl_2}=n_{H_2}=0,1\left(mol\right)\\b, m_{Zn\left(p.\text{ứ}\right)}=0,1.65=6,5\left(g\right)\\ n_{HCl}=0,1.2=0,2\left(mol\right)\\ m_{HCl}=0,2.36,5=7,3\left(g\right)\\ d,m_{ZnCl_2}=136.0,1=13,6\left(g\right)\)