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)

400ml = 0,4l

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

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

        1         2              1           1

      0,1                       0,1       0,1

a) \(n_{H2}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)

\(V_{H2\left(dktc\right)}=0,1.22,4=2,24\left(l\right)\)

\(n_{ZnCl2}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)

\(C_{M_{ZnCl2}}=\dfrac{0,1}{0,4}=0,25\left(M\right)\)

 Chúc bạn học tốt

\(a)Zn+2HCl\rightarrow ZnCl_2+H_2\\ b)n_{Zn}=\dfrac{6,5}{65}=0,1mol\\ Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,1      0,2             0,1             0,1

\(m_{ZnCl_2}=0,1.136=13,6g\\ c)V_{H_2}=0,1.24,79=2,479l\\ d)C_{\%HCl}=\dfrac{0,2.36,5}{200}\cdot100\%=3,65\%\)

\(a)n_{H_2}=\dfrac{6,1975}{24,79}=0,25mol\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{Zn}=n_{H_2}=0,25mol\\ m_{Zn}=0,25.65=16,25g\\ b)n_{HCl}=0,25.2=0,5mol\)

\(a)\) Theo định luật bảo toàn khối lượng:

\(m_{Zn}+m_{HCl}=m_{ZnCl_2}+m_{H_2}\\ b)\Rightarrow m_{H_2}=m_{Zn}+m_{HCl}-m_{ZnCl_2}\\ =6,5+7,3-13,6\\ =0,2g\)

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

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!

Zn+2Hcl->Zncl2+H2

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

n Zn=6,5\65=0,1 mol

=>m, ZnCl2=0,1.136=13,6g

=>VH2=0,1.22,4=2,24l

Sửa đề: Cho \(65g\) kẽm

\(a,PTHH:Zn+2HCl\to ZnCl_2+H_2\\ b,n_{H_2}=\dfrac{24,79}{24,79}=1(mol)\\ \Rightarrow m_{H_2}=1.2=2(g)\\ \text {Bảo toàn KL}:m_{HCl}=m_{ZnCl_2}+m_{H_2}-m_{Zn}=136+2-65=73(g)\)

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

Theo PT: \(n_{Zn}=n_{H_2}=n_{ZnCl_2}=0,2\left(mol\right)\)

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

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

Ta có \(n_{Zn}=\dfrac{13}{65}=0.2\left(mol\right)\)

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

           0,2---->0,4----->0,2------->0,2(mol)

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

\(m_{ZnCl_2}=0,2\cdot\left(65+35,5\cdot2\right)=27,2\left(g\right)\)

Tick nha