a, Ta có pt : \(Zn+2HCL->ZnCl_2+H_2\)

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

Theo pt (ở câu a ) , có :  \(n_{Zn}=n_{H_2}=0,1\left(mol\right)\)

\(V_{H_2\left(DKTC\right)}=0,1\times22,4=2,24\left(l\right)\)

n_Zn= \(\dfrac{6,5}{65}\) = 0,1 (mol)
a. PTHH: Zn + 2HCl -> ZnCl₂ + H₂

               0,1              -> 0,1 -> 0,1  (mol) 
b. n_H2 = 0,1 x 22,4 = 2,24 (l)
 

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

b, Ta có: \(n_{Zn}=\dfrac{32,5}{65}=0,5\left(mol\right)\)

Theo PT: \(n_{H_2}=n_{Zn}=0,5\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,5.22,4=11,2\left(l\right)\)

c, Theo PT: \(n_{ZnCl_2}=n_{Zn}=0,5\left(mol\right)\)

\(\Rightarrow m_{ZnCl_2}=0,5.136=68\left(g\right)\)

\(n_{Zn}=\dfrac{26}{65}=0.4\left(mol\right)\)

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

\(0.4.......0.8....................0.4\)

\(m_{HCl}=0.8\cdot36.5=29.2\left(g\right)\)

\(V_{H_2}=0.4\cdot22.4=8.96\left(l\right)\)

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

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)

`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)PTHH:`

`Zn + 2HCl -> ZnCl_2 + H_2`

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

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

`b)m_[HCl]=0,2.36,5=7,3(g)`

`c)`

`@m_[ZnCl_2]=0,1.136=13,6(g)`

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

\(n_{Zn}=\dfrac{9.75}{65}=0.15\left(mol\right)\)

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

\(0.15.....0.3...................0.15\)

\(V_{H_2}=0.15\cdot22.4=3.36\left(l\right)\)

\(m_{HCl}=0.3\cdot36.5=10.95\left(g\right)\)

a. Zn + 2HCl → ZnCl₂ + H₂

b. n_Zn = n\(_{ZnCl_2}\) =\(\dfrac{13}{65}=0,2\left(mol\right)\) => m\(_{ZnCl_2}\)= 0,2.136 = 27,2(g)

c. n\(_{H_2}\)= n_Zn = 0,2 (mol) => V\(_{H_2}\)=0,2.22,4 = 4,48 (lít)

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a)

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

b),c)

Theo PTHH :

\(n_{ZnCl_2} = n_{H_2} = n_{Zn} = \dfrac{13}{65} = 0,2(mol)\)

Vậy : 

\(m_{ZnCl_2} = 0,2.136 = 27,2(gam)\\ V_{H_2} =0,2.22,4 = 4,48(lít)\)

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

b+c)

Ta có: \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)=n_{ZnCl_2}=n_{H_2}\)

\(\Rightarrow\left\{{}\begin{matrix}m_{ZnCl_2}=0,2\cdot136=27,2\left(g\right)\\V_{H_2}=0,2\cdot22,4=4,48\left(l\right)\end{matrix}\right.\)