\(n_{Zn}=\dfrac{9.75}{65}=0.15\left(mol\right)\)

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

\(0.15.....0.3...................0.15\)

\(V_{H_2}=0.15\cdot22.4=3.36\left(l\right)\)

\(m_{HCl}=0.3\cdot36.5=10.95\left(g\right)\)

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

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

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

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

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

n H₂ = \(\dfrac{6,72}{22,4}=0,3\) (mol)

=> n Zn = n H₂ = 0,3 mol

=> m Zn = 0,3.65= 19,5 (g)

\(n_{HCl}=0,2.2=0,4\left(mol\right)\\ a.Zn+2HCl\rightarrow ZnCl_2+H_2\\ b.0,2........0,4.......0,2.......0,2\left(mol\right)\\ m=m_{Zn}=0,2.65=13\left(g\right)\\ c.m_{ZnCl_2}=136.0,2=27,2\left(g\right)\\ d.V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\)

`a)PTHH:`

`Zn + 2HCl -> ZnCl_2 + H_2`

`0,1`  `0,2`               `0,1`     `0,1`          `(mol)`

`n_[Zn]=[6,5]/65=0,1(mol)`

`b)m_[HCl]=0,2.36,5=7,3(g)`

`c)`

`@m_[ZnCl_2]=0,1.136=13,6(g)`

`@V_[H_2]=0,1.22,4=2,24(l)`

\(a.Mg+2HCl\rightarrow MgCl_2+H_2\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ b.Đặt:\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\\ n_{H_2}=\dfrac{18,48}{22,4}=0,825\left(mol\right)\\ Tacó:\left\{{}\begin{matrix}24x+27y=17,1\\x+\dfrac{3}{2}y=0.825\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0,375\\y=0,3\end{matrix}\right.\\ \Rightarrow\%m_{Mg}=\dfrac{0,375.24}{17,1}.100=52,63\%\\ \%m_{Al}=47,37\%\)

A) Phương trình phản ứng:

$Zn+2HCl\rightarrow ZnCl_2+H_2$

B) Ta có:

$n_{Zn}=\dfrac{13}{65}=0,2\ mol$

Theo phương trình:

$n_{HCl}=2n_{Zn}=0,4\ mol$

Đổi $300\ ml=0,3\ lít$

$C_M(HCl)=\dfrac{0,4}{0,3}=\dfrac{4}{3}\approx1,33\ M$

C) Theo phương trình:

$n_{H_2}=n_{Zn}=0,2\ mol$

$V_{H_2}=0,2\cdot22,4=4,48\ lít$

D) Phương trình phản ứng:

$CuO+H_2\xrightarrow{t^\circ}Cu+H_2O$

Ta có: $n_{CuO}=\dfrac{12}{80}=0,15\ mol$

$n_{H_2}=0,2\ mol$

Theo phương trình, tỉ lệ mol $CuO:H_2=1:1$.

Vì $0,2>0,15$ nên $H_2$ dư.

$n_{H_2\ dư}=0,2-0,15=0,05\ mol$

Vậy $H_2$ dư $0,05\ mol$, còn $CuO$ phản ứng hết.