Theo gt ta có: $n_{Zn}=0,1(mol);n_{CuO}=0,25(mol)$

a, $Zn+2HCl\rightarrow ZnCl_2+H_2$

$CuO+H_2\rightarrow Cu+H_2O$

b, Ta có: $n_{ZnCl_2}=0,1(mol)\Rightarrow m_{ZnCl_2}=13,6(g)$

b, Ta có: $n_{H_2}=0,1(mol)$ 

Sau phản ứng chất còn dư là CuO dư 0,15 mol

$\Rightarrow m_{CuO/du}=12(g)$

a)

$Zn + 2HCl \to ZnCl_2 + H_2$
$CuO + H_2 \xrightarrow{t^o} Cu + H_2O$

b)

n ZnCl2 = n Zn = 6,5/65 = 0,1(mol)

=> m ZnCl2 = 0,2.136 = 13,6(gam)

c) n H2 = n Zn = 0,1 mol

CuO + H2 --to--> Cu + H2O

n CuO = 20/80 = 0,25 > n H2 = 0,1 nên CuO dư

n CuO pư = n H2 = 0,1 mol

=> m CuO dư = 20 - 0,1.80 = 12(gam)

n_Zn  = 6,5 : 65 = 0,1 (mol) 
pthh: Zn+2HCl -> ZnCl₂ + H2 
        0,1   0,1                      0,1
=> m_HCl  = 0,1 . 36,5 = 3,65(g) 
pthh : CuO + H₂ -to-> Cu + H₂O 
                     0,1          0,1 
=> m_Cu  = 0,1 . 64 = 6,4 (g) 
 

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

Ta có: \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)=n_{ZnCl_2}=n_{H_2}\)

\(\Rightarrow\left\{{}\begin{matrix}m_{ZnCl_2}=0,1\cdot136=13,6\left(g\right)\\V_{H_2}=0,1\cdot22,4=2,24\left(l\right)\end{matrix}\right.\)

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

Theo PTHH: \(n_{H_2}=n_{Cu}=0,1\left(mol\right)\)

\(\Rightarrow m_{Cu}=0,1\cdot64=6,4\left(g\right)\)

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

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

Theo PT: \(n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,1\left(mol\right)\)

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

c, Ta có: \(m_{ZnCl_2}=0,1.136=13,6\left(g\right)\)

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

Theo PT: \(n_{Cu}=n_{H_2}=0,1\left(mol\right)\)

\(\Rightarrow m_{Cu}=0,1.64=6,4\left(g\right)\)

Bạn tham khảo nhé!

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

\(n_{HCl}=0,2.1=0,2\left(mol\right)\)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

Xét tỉ lệ:  \(\dfrac{0,2}{1}>\dfrac{0,2}{2}\) => Zn dư, HCl hết

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

__________0,2-------------->0,1

=> V_H2 = 0,1.22,4 = 2,24(l)

b)

PTHH: 2H₂ + O₂ --t^o--> 2H₂O

______0,1->0,05

=> m_O2 = 0,05.22,4 = 1,12 (l)

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

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

______0,2_________________0,2 (mol)

b, V_H2 = 0,2.22,4 = 4,48 (l)

c, Ta có: \(n_{FeO}=\dfrac{7,2}{72}=0,1\left(mol\right)\)

PT: \(FeO+H_2\underrightarrow{t^o}Fe+H_2O\)

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

Theo PT: \(n_{Fe}=n_{FeO}=0,1\left(mol\right)\)

⇒ m_Fe = 0,1.56 = 5,6 (g)

Bạn tham khảo nhé!

a) Zn  + 2HCl \(\rightarrow\)ZnCl₂ + H₂

b) m_Zn = \(\dfrac{13}{65}\)=0,2 (mol)

         Zn  + 2HCl \(\rightarrow\)ZnCl₂ + H₂

(mol) 0,2 ----------------------> 0,2

\(V_{H_2}\)= 0,2 . 22,4 = 4,48(lít)

c)\(n_{FeO}\)=\(\dfrac{7,2}{72}\)=0,1 (mol)

         H₂ + FeO \(\underrightarrow{t^o}\)Fe + H₂O

(mol)         0,1----->0,1

m_Fe = 0,1 . 56 = 5,6(g)

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

\(PTHH:Zn+2HCl->ZnCl_2+H_2\)

ap dung DLBTKL ta co

\(m_{Zn}+m_{HCl}=m_{ZnCl_2}+m_{H_2}\)

\(=>m_{H_2}=m_{Zn}+m_{HCl}-m_{ZnCl_2}\\ =>m_{H_2}=13+14,6-27,2\\ =>m_{H_2}=0,4\left(g\right)\)

\($a/$\\ Zn+2HCl \to ZnCl_2+H_2\\ b/\\ n_{Zn}=0,1(mol)\\ n_{HCl}=0,1.2=0,2(mol)\\ m_{ddHCl}=\frac{0,2.36,5.100}{7,3}=100(g)\\ C\%_{ZnCl_2}=\frac{0,1.136}{100+6,5-0,1.2}.100\%=12,8\%\)

\(a/\\ Zn+2HCl \to ZnCl_2+H_2\\ b/\\ n_{Zn}=\frac{6,5}{65}=0,1(mol)\\ n_{HCl}=0,2(mol)\\ m_{ddsaupu}=\frac{0,1.36,5.100}{7,3}+6,5-0,1.2=56,3(g)\\ C\%_{ZnCl_2}=\frac{0,1.136}{56,3}.100\%=24,16\%\)

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

PTHH: Zn + 2HCl → ZnCl₂ + H₂

Mol:     0,1    0,05       0,1

b, \(m_{HCl}=0,05.36,5=2,92\left(g\right)\)

  \(m_{ddHCl}=\dfrac{2,92}{7,3\%}=40\left(g\right)\)

  \(m_{ddsaupứ}=6,5+40-0,1.2=46,3\left(g\right)\)

\(\Rightarrow C\%_{ddZnCl_2}=\dfrac{0,1.136.100\%}{46,3}=29,4\%\)