a. \(n_{Zn}=\dfrac{6.5}{65}=0,1\left(mol\right)\)

PTHH : Zn + 2HCl -> ZnCl₂ + H₂

            0,1      0,2                  0,1

b. \(V_{H_2}=0,1.22,4=2,24\left(l\right)\)

c. \(m_{HCl}=0,2.36,5=7,3\left(g\right)\)

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

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

0,1      0,2                       0,1

\(V_{H_2}=0,1\cdot22,4=2,24l\)

\(m_{HCl}=0,2\cdot36,5=7,3g\)

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

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

   0,05    0,1          0,05       0,05

b)\(m_{ZnCl_2}=0,05\cdot136=6,8g\)

c)\(V_{H_2}=0,05\cdot22,4=1,12l\)

\(a.Zn+2HCl\rightarrow ZnCl_2+H_2\\ b.n_{Zn}=\dfrac{65}{65}=1\left(mol\right)\\ n_{ZnCl_2}=n_{Zn}=1\left(mol\right)\\ \Rightarrow m_{ZnCl_2}=1.136=136\left(g\right)\\ c.n_{H_2}=n_{Zn}=1\left(mol\right)\\ \Rightarrow V_{H_2}=1.22,4=22,4\left(l\right)\)

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

Áp dung định luật BTKL : 

\(m_{H_2}=13+14.6-27.2=0.4\left(g\right)\)

\(n_{H_2}=\dfrac{0.4}{2}=0.2\left(mol\right)\)

\(V_{H_2}=0.2\cdot22.4=4.48\left(l\right)\)

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

Áp dụng định luật BTKL : 

\(m_{Mg}+m_{HCl}=m_{MgCl_2}+m_{H_2}\)

\(\Rightarrow m_{Mg}=9.5+0.2-7.3=2.4\left(g\right)\)

\(n_{H_2}=\dfrac{0.2}{2}=0.1\left(mol\right)\)

\(V_{H_2}=0.1\cdot22.4=2.24\left(l\right)\)

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

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+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!

\(a.Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{Zn}=n_{ZnCl_2}=n_{H_2}=0,25\left(mol\right)\\ \Rightarrow m_{Zn}=0,25.65=16,25\left(g\right)\\ m_{ZnCl_2}=0,25.136=34\left(g\right)\\ b.FeO+H_2-^{t^o}\rightarrow Fe+H_2O\\ Tacó:n_{Fe}=n_{H_2}=0,25\left(mol\right)\\ \Rightarrow m_{Fe}=0,25.56=14\left(g\right)\)

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

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

\(n_{Zn}=\frac{m}{M}=\frac{13}{65}=0,2mol\)

\(a)m_{ZnCl_2}=n.M=0,4.136=54,4g\)

\(b)V_{H_2}=n.22,4=0,2.22,4=4,48l\)

c)mk đã kê mol trên r đó bạn đó là số mol của HCl