1 cách cũng được

\(nZn=\dfrac{6,5}{65}=0,1mol\)

PTHH:

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

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

a, \(VH_2=0,1.22,4=2,24lít\)

b, \(mHCl=0,2.36,5=3,65g\)

c, cách 1:

   \(mZnCl_2=0,2.136=27,2gam\)   

 cách 2 :

\(mH_2=0,1.2=0,2gam\)

BTKL: mZn+mHCl=mZnCl2+H2

            6,5+3,65   =mZnCl2+0,2

sao cái cách 2 btkl mình cs tính kq mà s nhìn nó lạ lắm

a. \(n_{Zn}=\dfrac{19.5}{65}=0,3\left(mol\right)\)

\(n_{HCl}=\dfrac{14.6}{36.5}=0,4\left(mol\right)\)

PTHH : Zn + 2HCl -> ZnCl₂ + H₂  

                      0,4        0,2        0,2

Ta thấy : \(\dfrac{0.3}{1}>\dfrac{0.4}{2}\) => Zn dư , HCl đủ

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

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

a) Zn + 2HCl --> ZnCl₂ + H₂ ↑ (1)

             0,3   -->0,15  -->0,15       (mol)

n_Zn= 19,5/65 = 0,3 mol

n_HCl= 14,5/37,5 = 0,3 mol

Ta có : n_{Zn bài ra} / n_{Zn phương trình}=0,3/1=0,3 (mol)

n_{HCl bài ra} / n_{HCl phương trình}=0,3/2=0,15 (mol)

=> HCl đủ,Zn dư

b) Theo PT(1) => n_H2=0,15(mol)

=>V_H2=0,15 x 22,4 = 3,36(l)

c) Theo PT(1) => n_ZnCl2=0,15(mol)

=>m_ZnCl2=0,15 x 136 = 20,4(g)

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)

\(n_{Zn}=\dfrac{6.5}{65}=0.1\left(mol\right)\)

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

\(0.1.................................0.1\)

\(Đặt:n_{CuO\left(pư\right)}=x\left(mol\right)\)

\(CuO+H_2\underrightarrow{t^0}Cu+H_2O\)

\(x............x\)

\(m_{cr}=6-80x+64x=5.2\left(g\right)\)

\(\Rightarrow x=0.05\)

\(H\%=\dfrac{0.05}{0.075}\cdot100\%=66.67\%\)

PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

___0,3____0,6_____0,3____0,3 (mol)

a, \(m_{Zn}=0,3.65=19,5\left(g\right)\)

b, \(m_{HCl}=0,6.36,5=21,9\left(g\right)\)

c, \(m_{ZnCl_2}=0,3.136=40,8\left(g\right)\)

Bạn tham khảo nhé!

Zn+2HCl->ZnCl₂+H₂

0,3---0,6-----0,3----0,3 mol

=>m _Zn=0,3.65=19,5g

=>m _HCl=0,6.35,6=21,9g

=>m _ZnCl2=0,3.136=40,8g

Zn +2 HCl ---> ZnCl₂ + H₂

0,1-----0,2----------0,1-------------0,1 mol

ZnO + 2HCl ---> ZnCl₂ + H₂O

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

n _H2=\(\dfrac{2,24}{22,4}=0,1mol\)

=>m _Zn=0,1.65=6,5g

=>m _HCl(1)=0,2.36,5=7,3g

=>m _HCl(2)=14,6g -> nHCl=0,4 mol

=>%m _Zn=\(\dfrac{6,5}{6,5+14,4}.100=31,1\%\)

=>%m _ZnO=68,9%

b)

->m _HCl=0,6.36,5=21,9g

->m _ZnCl2=0,3.136=40,8g

Lần đầu thấy H2OO

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

b, \(n_{Zn}=\dfrac{39}{65}=0,6\left(mol\right)\)

Theo PT: \(n_{HCl}=2n_{Zn}=1,2\left(mol\right)\Rightarrow m_{HCl}=1,2.36,5=43,8\left(g\right)\)

c, \(n_{H_2}=n_{Zn}=0,6\left(mol\right)\Rightarrow V_{H_2}=0,6.24,79=14,874\left(l\right)\)

d, - Quỳ tím hóa đỏ do HCl dư.

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

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