\(m_{ct}=\dfrac{10.200}{100}=20\left(g\right)\)

\(n_{NaOH}=\dfrac{20}{40}=0,5\left(mol\right)\)

a) Pt : \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O|\)

                  2              1                1               2

                 0,5           0,25          0,25

\(n_{H2SO4}=\dfrac{0,5.1}{2}=0,25\left(mol\right)\)

\(m_{H2SO4}=0,25.98=24,5\left(g\right)\)

\(m_{ddH2SO4}=\dfrac{24,5.100}{20}=122,5\left(g\right)\)

b) \(n_{Na2SO4}=\dfrac{0,25.1}{1}=0,25\left(mol\right)\)

⇒ \(m_{Na2SO4}=0,25.142=35,5\left(g\right)\)

\(m_{ddspu}=200+122,5=322,5\left(g\right)\)

\(C_{Na2SO4}=\dfrac{35,5.100}{322,5}=11\)⁰/₀

 Chúc bạn học tốt

\(n_{H_2}=\dfrac{10.08}{22.4}=0.45\left(mol\right)\)

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

\(0.3.............................0.3..........0.45\)

\(m_{Al}=0.3\cdot27=8.1\left(g\right)\)

\(m_{Al_2\left(SO_4\right)_3}=0.3\cdot342=102.6\left(g\right)\)

\(n_{Fe_2O_3}=\dfrac{8}{160}=0,05\left(mol\right)\)

PTHH: Fe₂O₃ + 3H₂SO₄ --> Fe₂(SO₄)₃ + 3H₂O

______0,05------>0,15--------->0,05

=> m_H2SO4 = 0,15.98 = 14,7(g)

=> \(C\%\left(H_2SO_4\right)=\dfrac{14,7}{100}.100\%=14,7\%\)

\(C\%\left(Fe_2\left(SO_4\right)_3\right)=\dfrac{0,05.400}{8+100}.100\%=18,52\%\)

PTHH: Fe₂(SO₄)₃ + 6NaOH --> 2Fe(OH)₃\(\downarrow\) + 3Na₂SO₄

________0,05----------------------->0,1

=> mFe(OH)₃ = 0,1.107=10,7(g)

a)

$Zn + H_2SO_4 \to ZnSO_4 + H_2$
Theo PTHH : 

$n_{H_2} = n_{ZnSO_4} = n_{H_2SO_4} = n_{Zn} = \dfrac{21,6}{65} = 0,33(mol)$
$V_{H_2} = 0,33.22,4 = 7,392(lít)$
b)

$m_{ZnSO_4} = 0,33.161 = 53,13(gam)$
c)

$V_{H_2SO_4} = \dfrac{0,33}{2} = 0,165(lít)$

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

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

PTHH: Zn + H₂SO₄ ---> ZnSO₄ + H₂

         0,2--->0,2--------->0,2------>0,2

\(V_{H_2}=0,2.22,4=4,48\left(l\right)\\ m_{H_2SO_4}=\dfrac{0,2.98}{20\%}=98\left(g\right)\\ \rightarrow V_{ddH_2SO_4}=\dfrac{98}{1,14}=86\left(ml\right)=0,086\left(l\right)\\ \rightarrow C_{M\left(H_2SO_4\right)}=\dfrac{0,2}{0,086}=2,33M\)