\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(n_{Al}=\dfrac{3,375}{27}=0,125\)

\(\Rightarrow n_{H_2}=\dfrac{0,125}{2}.3=0,1875mol\)      \(\Rightarrow V_{H_2}=0,1875.22,4=4,2l\)

\(n_{AlCl_3}=n_{Al}=0,125mol\)       \(\Rightarrow m_{AlCl_3}=0,125.133,5=16,6875g\)

`Mg + 2HCl -> MgCl_2 + H_2`

`0,15`  `0,3`                           `0,15`       `(mol)`

`n_[Mg]=[3,6]/24=0,15(mol)`

`a)V_[H_2]=0,15.22,4=3,36(l)`

`b)m_[HCl]=0,3.36,5=10,95(g)`

`c)`

`H_2 + CuO` $\xrightarrow{t^o}$ `Cu + H_2 O`

`0,15`                  `0,15`                 `(mol)`

  `=>m_[Cu]=0,15.64=9,6(g)`

\(n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right)\)

PTHH: Mg + 2HCl ---> MgCl₂ + H₂

          0,15->0,3------------------>0,15

          CuO + H₂ --t^o--> Cu + H₂O

                    0,15------>0,15

=> \(V_{H_2}=0,15.22,4=3,36\left(l\right)\\ m_{HCl}=0,3.36,5=10,95\left(g\right)\\ m_{Cu}=0,15.64=9,6\left(g\right)\)

\(a) 2Al + 6HCl \to 2AlCl_3 + 3H_2\\ b)n_{AlCl_3} = n_{Al} = \dfrac{8,1}{27} = 0,3(mol)\\ \Rightarrow m_{AlCl_3} = 0,3.133,5 = 40,05(gam)\\ c) n_{HCl} = 3n_{Al} = 0,9(mol)\\ \Rightarrow C_{M_{HCl}} = \dfrac{0,9}{0,5} = 1,8M\)

nH2 = 4.48/22.4 = 0.2 (mol) 

Zn + 2HCl => ZnCl2 + H2 

0.2......0.4.....................0.2

mZn = 65*0.2 = 13 (g) 

mHCl = 0.4*36.5 = 14.6 (g) 

C%HCl = 14.6*100%/200 = 7.3%

\(n_{HCl}=\dfrac{18.25}{36.5}=0.5\left(mol\right)\)

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

\(b.\)

\(n_{Mg}=n_{H_2}=\dfrac{1}{2}\cdot n_{HCl}=\dfrac{1}{2}\cdot0.5=0.25\left(mol\right)\)

\(m_{Mg}=0.25\cdot24=6\left(g\right)\)

\(V_{H_2}=0.25\cdot22.4=5.6\left(l\right)\)

\(c.\)

\(V_{H_2\left(tt\right)}=5.6\cdot90\%=5.04\left(l\right)\)

\(n_{HCl}=0,3.1=0,3\left(mol\right)\\ pthh:Fe+2HCl\rightarrow FeCl_2+H_2\) 
          0,15     0,3                      0,15 
\(m_{Fe}=0,15.56=8,4\left(g\right)\\ V_{H_2}=0,15.22,4=3,36\left(l\right)\)

a) Zn + 2HCl --> ZnCl₂ + H₂

b) \(n_{Zn}=\dfrac{1,3}{65}=0,02\left(mol\right)\)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂ 

           0,02-->0,04---------->0,02

=> V_H2 = 0,02.22,4 = 0,448 (l)

c) m_HCl = 0,04.36,5 = 1,46 (g)

\(a.Mg+2HCl->MgCl_2+H_2\\ MgO+2HCl->MgCl_2+H_2O\\ b.n_{H_2}=\dfrac{2,24}{22,4}=n_{Mg}=0,1mol\\ \%m_{Mg}=\dfrac{0,1.24}{6}=40\%;\%m_{MgO}=60\%\\ n_{MgO}=\dfrac{0,6.6}{40}=0,09\left(mol\right)\\ n_{MgCl_2}=0,1+0,09=0,19\left(mol\right)\\ n_{HCl}=0,19.2=0,38\left(mol\right)\\ V_{ddHCl}=\dfrac{0,38.36,5}{0,2.1,1}=63,0\left(mL\right)\\ C_{M\left(MgCl_2\right)}=\dfrac{0,19}{0,063}=3,0\left(M\right)\)

a) $n_{CaCO_3} = 0,15(mol)$
$CaCO_3 + 2HCl \to CaCl_2 + CO_2 + H_2O$
$n_{HCl} = 2n_{CaCO_3} = 0,3(mol)$
$m_{dd\ HCl} = \dfrac{0,3.36,5}{7,3\%} = 150(gam)$

b)

$n_{CaCl_2} = n_{CO_2} = n_{CaCO_3} =0,15(mol)$
$V_{CO_2} = 0,15.22,4 = 3,36(lít)$
c)

$m_{dd} = 15 + 150 - 0,15.44 = 158,4(gam)$

$C\%_{CaCl_2} = \dfrac{0,15.111}{158,4}.100\% = 10,51\%$