\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right);n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\a, Zn+2HCl\rightarrow ZnCl_2+H_2\\ b,V\text{ì}:\dfrac{0,2}{1}>\dfrac{0,1}{1}\Rightarrow Zn\text{dư}\\ \Rightarrow n_{Zn\left(p.\text{ứ}\right)}=n_{ZnCl_2}=n_{H_2}=0,1\left(mol\right)\\b, m_{Zn\left(p.\text{ứ}\right)}=0,1.65=6,5\left(g\right)\\ n_{HCl}=0,1.2=0,2\left(mol\right)\\ m_{HCl}=0,2.36,5=7,3\left(g\right)\\ d,m_{ZnCl_2}=136.0,1=13,6\left(g\right)\)

a, Bảo toàn e:

\(2n_{Zn}=2n_{H_2}\)

\(\Leftrightarrow n_{Zn}=n_{H_2}=0,3\left(mol\right)\)

\(\Rightarrow m_{Zn}=19,5\left(g\right)\)

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

b, Bảo toàn nguyên tố H:

\(n_{HCl}=2n_{H_2}=0,6\left(mol\right)\)

\(\Rightarrow m_{HCl}=21,9\left(g\right)\)

a) Zn + 2HCl --> ZnCl₂ + H₂

b) \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

_____0,15<-------------------0,15

=> m_Zn = 0,15.65 = 9,75(g)

c) \(n_{HCl}=\dfrac{3,65}{36,5}=0,1\left(mol\right)\)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

__________0,1------------->0,05

=> V_H2 = 0,05.22,4 = 1,12(l)

\(n_{HCl}=0,25.2=0,5\left(mol\right)\\ a,PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ b,n_{Zn}=n_{H_2}=n_{ZnCl_2}=\dfrac{0,5}{2}=0,25\left(mol\right)\\ m_{Zn}=0,25.65=16,25\left(g\right)\\ c,V_{H_2\left(đktc\right)}=0,25.22,4=5,6\left(l\right)\)

Không biết đúng không nữa;-;;;

a) PTHH: Zn + 2HCl -> ZnCl2 + H2 

b) HCl=250ml=0,25l

n2HCl= V/22,4= 0,5/22,4= 0,02(mol)

Zn  +  2HCl -> ZnCl2  +  H2

1          2              1         1

0,01 <-0,5--------------> 0,01

mZn= n.M= 0,01.65= 0,65(gam)

c) VH2=n . 22,4= 0,01 . 22,4= 0,224(l)
 

\(a,PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ Áp.dụng.ĐLBTKL,ta.có:\\ m_{Zn}+m_{HCl}=m_{ZnCl_2}+m_{H_2}\\ m_{H_2}=m_{Zn}+m_{HCl}-m_{ZnCl_2}=6,5+7,3-13,6=0,2\left(g\right)\)

a) 

Zn  +  2HCl  →  ZnCl₂   +  H₂ 

b) nZn = \(\dfrac{3,5}{65}\)=\(\dfrac{7}{130}\) mol 

Theo tỉ lệ phản ứng => nH₂ = nZn= \(\dfrac{7}{130}\)mol

<=> V H₂ = \(\dfrac{7}{130}\).22,4 = 1,206 lít

c) nZnCl₂ = nZn => mZnCl₂ = \(\dfrac{7}{130}\).136= 7,32 gam

B

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)