\(n_{Mg}=\dfrac{10,8}{24}=0,45\left(mol\right)\\ pthh:Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\) 
           0,45     0,45                          0,45 
\(m_{H_2SO_4}=0,45.98=44,1\left(g\right)\\ C\%_{H_2SO_4}=\dfrac{44,1}{176,4}.100\%=25\%\\ V_{H_2}=0,45.22,4=10,08\left(l\right)\)

Ta có: \(n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\)

PT: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)

______0,2_____0,4_____0,2 (mol)

a, \(m_{CuCl_2}=0,2.135=27\left(g\right)\)

b, \(m_{HCl}=0,4.36,5=14,6\left(g\right)\Rightarrow C\%_{HCl}=\dfrac{14,6}{300}.100\%\approx4,867\%\)

c, Ta có: m _{dd sau pư} = 16 + 300 = 316 (g)

\(\Rightarrow C\%_{CuCl_2}=\dfrac{27}{316}.100\%\approx8,54\%\)

`Fe_2O_3+3H_2SO_4->Fe_2(SO_4)_3+3H_2O`

0,0625----------0,1875---------0,0625 mol

`->n_(Fe_2O_3)=10/160=0,0625mol`

`->m_(Fe_2(SO_4)_3)=0,0625.400=25g`

`->C%(H_2SO_4)=((0,1875.98)/(450)).100%=4,083%`

`#YBtran<3`

\(n_{Fe_2O_3}=\dfrac{10}{160}=0,0625\left(mol\right)\\ Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\\ n_{Fe_2\left(SO_4\right)_3}=n_{Fe_2O_3}=0,0625\left(mol\right)\\ a,m=m_{Fe_2\left(SO_4\right)_3}=400.0,0625=25\left(g\right)\\ b,n_{H_2SO_4}=3.0,0625=0,1875\left(mol\right)\\ C\%_{ddH_2SO_4}=\dfrac{0,1875.98}{450}.100\%\approx4,083\%\)

\(n_{Al}=\dfrac{2,7}{27}=0,1mol\)

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

0,1       0,15             0 ,05            0,15

a)\(V=0,15\cdot22,4=3,36\left(l\right)\)

b)\(m_{H_2SO_4}=0,15\cdot98=14,7\left(g\right)\)

   \(\Rightarrow m_{ddHCl}=\dfrac{14,7}{9,8}\cdot100=150\left(g\right)\)

c) \(m_{H_2}=0,15\cdot2=0,3\left(g\right)\)

    \(m_{ddsau}=2,7+150-0,3=152,4\left(g\right)\)

    \(m_{Al_2\left(SO_4\right)_3}=0,05\cdot342=17,1\left(g\right)\)

   \(\Rightarrow C\%=\dfrac{17,1}{152,4}\cdot100=11,22\%\)

a) 

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

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

            0,2-->0,4----->0,2--->0,2

=> V_H2 = 0,2.22,4 = 4,48 (l)

b) m_HCl = 0,4.36,5 = 14,6 (g)

=> \(m_{dd.HCl}=\dfrac{14,6.100}{7,3}=200\left(g\right)\)

c)

m_{dd sau pư} = 13 + 200 - 0,2.2 = 212,6 (g)

m_ZnCl2 = 0,2.136 = 27,2 (g)

=> \(C\%=\dfrac{27,2}{212,6}.100\%=12,8\%\)

a)

$Fe + H_2SO_4 \to FeSO_4 + H_2$
$n_{H_2SO_4} = n_{H_2} = n_{Fe} = \dfrac{16,8}{56} = 0,3(mol)$

$V = 0,3.22,4 = 6,72(lít)$
$C_{M_{H_2SO_4}} = \dfrac{0,3}{0,25} = 1,2M$

b)

$n_{CuO} = \dfrac{16}{80} = 0,2(mol)$
$CuO + H_2 \xrightarrow{t^o} Cu + H_2O$
$n_{CuO} < n_{H_2}$ nên $H_2$ dư

$n_{Cu} = n_{CuO} = 0,2(mol)$
$m_{Cu} = 0,2.64 = 12,8(gam)$

\(n_{Fe}=\dfrac{m}{M}=\dfrac{16,8}{56}=0,3\left(mol\right)\\ PT:Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

       0,3     0,3                        0,3 (mol)

a) V= n. 22,4 = 0,3 . 22,4 = 6,72(l)

 \(C\%=\dfrac{m_{H_2SO_4}}{m_{dd}}.100\%=\dfrac{0,3.98}{200}.100\%=11,76\%\)

b) PT: \(H_2+CuO\underrightarrow{t^o}Cu+H_2O\)

         0,3                0,3 

=> m_Cu=n.M=0,3.64=19,2(g)

PTHH: \(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

Ta có: \(n_{CuO}=\dfrac{29,4}{80}=0,3675\left(mol\right)=n_{CuSO_4}=n_{H_2SO_4}\)

\(\Rightarrow\left\{{}\begin{matrix}m_{CuSO_4}=0,3675\cdot160=58,8\left(g\right)\\m_{H_2SO_4}=0,3675\cdot98=36,015\left(g\right)\\V_{H_2SO_4}=\dfrac{0,3675}{1}=0,3675\left(l\right)=367,5\left(ml\right)\end{matrix}\right.\)

\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\\ n_{HCl}=\dfrac{73.10\%}{36,5}=0,2\left(mol\right)\\ Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\\ Vì:\dfrac{0,1}{1}>\dfrac{0,2}{6}\\ \Rightarrow Fe_2O_3dư\\ a.Muối.tạo.thành:FeCl_3\\ n_{FeCl_3}=\dfrac{2}{6}.0,2=\dfrac{1}{15}\left(mol\right)\\ m_{FeCl_3}=\dfrac{1}{15}.162,5=\dfrac{65}{6}\left(g\right)\\ b.Chất.tan.ddA:FeCl_3\\ m_{ddFeCl_3}=m_{Fe_2O_3\left(p.ứ\right)}+m_{ddHCl}=\dfrac{1}{6}.0,2.160+73=\dfrac{235}{3}\left(g\right)\\ C\%_{ddFeCl_3}=\dfrac{\dfrac{65}{6}}{\dfrac{235}{3}}.100=13,83\%\)