a, Ta có: \(n_{Zn}=\dfrac{0,65}{65}=0,01\left(mol\right)\)

\(n_{HCl}=\dfrac{2,3}{36,5}=\dfrac{23}{365}\left(mol\right)\)

PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

Xét tỉ lệ: \(\dfrac{0,01}{1}< \dfrac{\dfrac{23}{365}}{2}\), ta được HCl dư.

Theo PT: \(n_{HCl\left(pư\right)}=2n_{Zn}=0,02\left(mol\right)\)

\(\Rightarrow n_{HCl\left(dư\right)}=\dfrac{23}{365}-0,02=\dfrac{157}{3650}\left(mol\right)\)

\(\Rightarrow m_{HCl\left(dư\right)}=\dfrac{157}{3650}.36,5=1,57\left(g\right)\)

b, Theo PT: \(n_{H_2}=n_{Zn}=0,01\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,01.22,4=0,224\left(l\right)\)

nZn=0.01 (mol)
nHCL=0.06 (mol)
pthh: Zn + 2HCL -> ZnCL2 +H2 
PT:     1         2            1          1
ĐB:     0.01    0.06       /           /
pứ:       0.01     0.02     0.01    0.01  
spu:      0           0.04     0.01     0.01  
a)vậy chất dư spu là HCL 
-> mHCL = 1.46 (g) 
b) V H2 đktc = 0.224 (L)

\(a.Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right);n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\\ LTL:\dfrac{0,2}{1}< \dfrac{0,5}{2}\Rightarrow HCldư\\ n_{HCl\left(pứ\right)}=2n_{Zn}=0,4\left(mol\right)\\\Rightarrow m_{HCl\left(dư\right)}=\left(0,5-0,4\right).36,5=3,65\left(g\right)\\ b.n_{ZnCl_2}=n_{Zn}=0,2\left(mol\right)\\ \Rightarrow m_{ZnCl_2}=0,2.136=27,2\left(g\right)\\ c.n_{H_2}=n_{Zn}=0,2\left(mol\right)\\ \Rightarrow V_{H_2}=0,2.22,4,=4,48\left(l\right)\\ d.3H_2+Fe_2O_3-^{t^o}\rightarrow2Fe+3H_2O \\ n_{Fe_2O_3}=\dfrac{19,2}{160}=0,12\left(mol\right)\\ LTL:\dfrac{0,2}{3}< \dfrac{0,12}{1}\Rightarrow Fe_2O_3dưsauphảnứng\\ \Rightarrow n_{Fe}=\dfrac{2}{3}n_{H_2}=\dfrac{2}{15}\left(mol\right)\\ \Rightarrow m_{Fe}=\dfrac{2}{15}.56=7,467\left(g\right)\)

a) n\(Zn\)=\(\dfrac{m}{M}\)=\(\dfrac{13}{65}\)=0,2(mol)

n\(HCl\)=\(\dfrac{m}{M}\)=\(\dfrac{18,25}{36,5}=\)0,5(mol)

PTHH : Zn + 2HCl->ZnCl\(2\) + H\(2\)

            0,2     0,5

Lập tỉ lệ mol : \(^{\dfrac{0,2}{1}}\)<\(\dfrac{0,5}{2}\)

n\(Zn\) hết , n\(HCl\) dư

-->Tính theo số mol hết

               Zn + 2HCl->ZnCl\(2\) + H\(2\)

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

n\(HCl\) dư= n\(HCl\)(đề) - n\(HCl\)(pt)= 0,5 - 0,4 = 0,1(mol)

m​\(HCl\) dư= 0,1.36,5 = 3,65(g)

b) m\(ZnCl2\) = n.M= 0,2.136= 27,2 (g)

c)V\(H2\)=n.22,4=0,2.22,4=4,48(l)

d) n\(Fe\)\(2\)O\(3\)=\(\dfrac{m}{M}\)=\(\dfrac{19,2}{160}\)=0,12 (mol)

 3H2 +Fe2O3 → 2Fe + 3H2O

  0,2       0,12

Lập tỉ lệ mol: \(\dfrac{0,2}{3}\)<\(\dfrac{0,12}{1}\)

    nH2 hết .Tính theo số mol hết

\(HCl\)

​ 3H2 +Fe2O3 → 2Fe + 3H2O

0,2->                  0,2

m\(Fe\)=n.M= 0,2.56= 11,2(g)

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

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

\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=1\left(mol\right)\\n_{H_2}=0,5\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{ddHCl}=\dfrac{1\cdot36,5}{20\%}=182,5\left(g\right)\\V_{H_2}=0,5\cdot22,4=11,2\left(l\right)\end{matrix}\right.\)

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

PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

____0,2____________0,2____0,2 (mol)

\(V_{H_2}=0,2.22,4=4,48\left(l\right)\)

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

c, \(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)

PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,2}{1}\), ta được H₂ dư.

Theo PT: \(n_{Cu}=n_{CuO}=0,1\left(mol\right)\Rightarrow m_{cr}=m_{Cu}=0,1.64=6,4\left(g\right)\)

$a\big)$

$n_{Zn}=\dfrac{3,25}{65}=0,05(mol)$

$Zn+2HCl\to ZnCl_2+H_2$

Theo PT: $n_{ZnCl_2}=n_{Zn}=0,05(mol)$

$\to m_{ZnCl_2}=0,05.136=6,8(g)$

$b\big)$

Theo PT: $n_{HCl}=2n_{Zn}=0,1(mol)$

$\to V_{dd\,HCl}=\dfrac{0,1}{0,5}=0,2(l)=200(ml)$

`Zn + 2HCl -> ZnCl_2 + H_2`

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

`n_[Zn]=[3,25]/65=0,05 (mol)`

`a)m_[ZnCl_2]=0,05.136=6,8(g)`

`b)V_[dd HCl]=[0,1]/[0,5]=0,2(M)`

a) Zn + 2HCl →ZnCl₂  + H₂

b) nZn = 6,5/65 = 0,1 mol . Theo tỉ lệ pư => nH₂ = nZn = nZnCl₂ =0,1 mol <=> V_H2(đktc) = 0,1.22,4 = 2,24 lít.

c) mZnCl₂ = 0,1 . 136 = 13,6 gam

d) nHCl =2nZn = 0,2 mol => mHCl = 0,2.36,5= 7,3 gam

Cách 2: áp dụng định luật BTKL => mHCl = mZnCl₂ + mH₂ - mZn 

<=> mHCl = 13,6 + 0,1.2 - 6,5 = 7,3 gam

\(n_{Fe}=\dfrac{11,2}{56}=0,2mol\)

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

0,2     0,4          0,2         0,2

\(V_{H_2}=0,2\cdot22,4=4,48l\)

\(m_{FeCl_2}=0,2\cdot\left(56+2\cdot35,5\right)=25,4g\)

\(a.Zn+2HCl\rightarrow ZnCl_2+H_2\\ b.n_{Zn}=n_{ZnCl_2}=0,1\left(mol\right)\\ \Rightarrow m_{ZnCl_2}=0,1.136=13,6\left(g\right)\\ c.n_{H_2}=n_{Zn}=0,1\left(mol\right)\\ \Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)