\(n_{Zn}=\dfrac{13}{65}=0,2(mol)\\ a,PTHH:Zn+2HCl\to ZnCl_2+H_2\\ b,n_{HCl}=2n_{Zn}=0,4(mol)\\ \Rightarrow m_{HCl}=0,4.36,5=14,6(g)\\ c,n_{H_2}=n_{Zn}=0,2(mol)\\ \Rightarrow V_{H_2}=0,2.22,4=4,48(l)\)

b) mHCl = 14,6 (g)

V H2 = 4,48 (l)

Giải thích các bước:

a) PTHH:  Zn +  2HCl → ZnCl2 + H2↑

b) nZn = 13 : 65 = 0,2 mol

Theo PTHH: nHCl = 2.nZn = 0,4 mol

mHCl = 0,4 . 36,5 = 14,6(g)

c) nH2 = nZn = 0,2 mol

VH2 = 0,2 . 22,4 = 4,48 (l)

PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)

a+b+c) Ta có: \(n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\)

\(\Rightarrow n_{H_2}=0,25\left(mol\right)=n_{MgCl_2}\) \(\Rightarrow\left\{{}\begin{matrix}m_{MgCl_2}=0,25\cdot95=23,75\left(g\right)\\V_{H_2}=0,25\cdot22,4=5,6\left(l\right)\end{matrix}\right.\)

d) PTHH: \(CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\)

Theo PTHH: \(n_{Cu}=n_{H_2}=0,25\left(mol\right)\) \(\Rightarrow m_{Cu}=0,25\cdot64=16\left(g\right)\)

a) Zn + 2HCl --> ZnCl₂ + H₂

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

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

           0,2-->0,4------------>0,2

=> V_H2 = 0,2.22,4 = 4,48 (l)

c) m_HCl = 0,4.36,5 = 14,6 (g)

nZn = 13 : 65 = 0,2 (mol) 
pthh : Zn + 2HCl --> ZnCl2 + H2
          0,2 ------------> 0,2 ---->0,2 (mol)  
VH2 = 0,2 . 22,4 = 4,48 (l) 
mZnCl2 = 0,2 . 36,5 = 14,6 (g) 

a) Zn + 2HCl --> ZnCl₂ + H₂

b) \(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

            0,3------------->0,3--->0,3

=> m_ZnCl2 = 0,3.136 = 40,8 (g)

c) V_H2 = 0,3.22,4 = 6,72 (l)

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

b. \(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\)

Mol theo PTHH : \(1:2:1:1\)

Mol theo phản ứng : \(0,3\rightarrow0,6\rightarrow0,3\rightarrow0,3\)

\(\Rightarrow m_{ZnCl_2}=n_{ZnCl_2}.M_{ZnCl_2}=0,3.\left(65+71\right)=40,8\left(g\right)\)

c. Từ b. \(\Rightarrow n_{H_2}=0,3\left(mol\right)\)

\(\Rightarrow V_{H_2}=n_{H_2}.22,4=0,3.22,4=6,72\left(l\right)\)

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

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!

a) Zn + 2HCl --> ZnCl₂ + H₂

b) \(n_{ZnCl_2}=\dfrac{13,6}{136}=0,1\left(mol\right)\)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

           0,1<-----------0,1----->0,1

=> m_Zn = 0,1.65 = 6,5 (g)

b) V_H2 = 0,1.22,4 = 2,24 (l)

`a)PTHH:`

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

`0,4`   `0,8`                          `0,4`   `(mol)`

`n_[Zn]=26/65=0,4(mol)`

`b)m_[HCl]=0,8.36,5=29,2(g)`

`c)V_[H_2]=0,4.22,4=8,96(l)`

a) \(n_{Zn}=\dfrac{16,25}{65}=0,25\left(mol\right)\)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

          0,25-------------------->0,25

=> V_H2 = 0,25.22,4 = 5,6 (l)

b)

PTHH: CuO + H₂ --t^o--> Cu + H₂O

                      0,25--->0,25

=> m_Cu = 0,25.64 = 16 (g)

\(a) Zn + H_2SO_4 \to ZnSO_4 + H_2\\ b) n_{H_2} = n_{Zn} = \dfrac{13}{65}=0,2(mol)\\ V_{H_2} = 0,2.22,4 = 4,48(lít)\\ c) CuO + H_2 \xrightarrow{t^o} Cu + H_2O\\ n_{CuO} = n_{H_2} = 0,2(mol)\\ m_{CuO} = 0,2.80 = 16(gam)\)

a)

    \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

a,PTHH: Zn+H2SO4-----ZnSO4 +H2

nZn = 0,2 mol

=> VH2=4,48 l

CuO + H2 ----- Cu +H2O

nH2= 0,2 mol 

=> mCuO= 16g