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

\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

0,2     0,2            0,2           0,2

a)\(V_{H_2}=0,2\cdot22,4=4,48l\)

b)\(m_{ZnSO_4}=0,2\cdot161=32,2g\)

  \(m_{ddZnSO_4}=30+200-0,2\cdot2=229,6g\)

  \(C\%=\dfrac{m_{ct}}{m_{dd}}\cdot100\%=\dfrac{32,2}{229,6}\cdot100\%=14,02\%\)

c)\(n_{CuO}=\dfrac{24}{80}=0,3mol\)

   \(CuO+H_2\rightarrow Cu+H_2O\)

   0,3        0,2     0,2

   \(m_{rắn}=m_{Cu}=0,2\cdot64=12,8g\)

0,2     0,2            0,2           0,2

a)

b)

c)

   0,3        0,2     0,2

`a)PTHH:`

`Mg + 2HCl -> MgCl_2 + H_2`

`0,3`      `0,6`             `0,3`       `0,3`    `(mol)`

`b) n_[Mg] = [ 7,2 ] / 24 = 0,3 (mol)`

 `=> V_[H_2] = 0,2 . 22,4 =6,72 (l)`

`c) C_[M_[HCl]] = [ 0,6 ] / [ 0,3 ] = 2 (M)`

nFe = 5.6/56 = 0.1 (mol) 

nHCl = 0.2*2 = 0.4 (mol) 

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

LTL : 0.1/1 < 0.4/2 => HCl dư 

mHCl dư = ( 0.4 - 0.2 ) * 36.5 = 7.3 (g) 

VH2 = 0.2*22.4 = 4.48 (l) 

CM FeCl2 = 0.1/0.2 = 0.5(M)

CM HCl dư = 0.2 / 0.2 = 1(M) 

\(n_{Al} = \dfrac{10,8}{27} = 0,4(mol)\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\\ n_{AlCl_3} = n_{Al} = 0,4(mol)\\ m_{AlCl_3} = 0,4.133,5 = 53,4(gam)\\ n_{HCl} =3 n_{Al} = 1,2(mol)\\ C\%_{HCl}= \dfrac{1,2.36,5}{200}.100\% = 21,9\%\)

a, \(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)

b, \(n_{CaCO_3}=\dfrac{10}{100}=0,1\left(mol\right)\Rightarrow n_{CO_2}=n_{CaCO_3}=0,1\left(mol\right)\)

\(\Rightarrow V_{CO_2}=0,1.22,4=2,24\left(l\right)\)

c, \(n_{HCl}=2n_{CaCO_3}=0,2\left(mol\right)\Rightarrow C_{M_{HCl}}=\dfrac{0,2}{0,1}=2\left(M\right)\)

d, \(m_{NaOH}=550.10\%=55\left(g\right)\Rightarrow n_{NaOH}=\dfrac{55}{40}=1,375\left(mol\right)\)

\(\Rightarrow\dfrac{n_{NaOH}}{n_{CO_2}}=\dfrac{1,375}{0,1}=13,75>2\)

→ Pư tạo muối trung hòa Na₂CO₃.

PT: \(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)

\(n_{Na_2CO_3}=n_{CO_2}=0,1\left(mol\right)\Rightarrow m_{Na_2CO_3}=0,1.106=10,6\left(g\right)\)

Theo gt ta có: $n_{Zn}=0,06(mol)$

$Zn+2HCl\rightarrow ZnCl_2+H_2$

a, Ta có: $n_{HCl}=0,12(mol)\Rightarrow m_{ddHCl}=30(g)$

b, Ta có: $n_{H_2}=0,06(mol)\Rightarrow V_{H_2}=1,344(l)$

a)

n Zn = 3,9.65 = 0,06(mol)

Zn + 2HCl $\to$ ZnCl2 + H2

Theo PTHH :

n HCl = 2n Zn =0,12(mol)

=> mdd HCl = 0,12.36,5/14,6% = 30(gam)

n H2 = n Zn = 0,06(mol)
V H2 = 0,06.22,4 = 1,344 lít

`Fe + 2HCl -> FeCl_2 + H_2 \uparrow`

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

`n_[Fe]=[5,6]/56=0,1(mol)`

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

`b)C_[M_[HCl]]=[0,2]/[0,1]=2(M)`

a. \(n_{SO_3}=\dfrac{8}{80}=0,1\left(mol\right)\)

\(C_M=\dfrac{0,1}{200}=0,0005\left(mol\text{/}l\right)\) 

câu 1 
\(n_{Fe}=\dfrac{14}{56}=0,25\left(mol\right)\\ pthh:Fe+2HCl\rightarrow FeCl_2+H_2\) 
           0,25   0,5         0,25       0,25 
\(m_{FeCl_2}=0,25.127=31,75g\\ V_{H_2}=0,25.22,4=5,6\\ C_{M\left(HCl\right)}=\dfrac{0,5}{0,2}=2,5M\)
câu 2 
1 ) \(m_{\text{dd}}=35+100=135g\\ 2,C\%=\dfrac{204}{204+100}.100=60\%\\ =>m\text{dd}=\dfrac{100.204}{60}=340g\)