\(n_{Al}=\dfrac{21,6}{27}=0,8\left(mol\right)\)

Pt : \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O|\)

        2             3                  1                3

       0,8          1,2               0,4             1,2

a) \(n_{H2}=\dfrac{0,8.3}{2}=1,2\left(mol\right)\)

\(V_{H2\left(dktc\right)}=1,2.22,4=26,88\left(l\right)\)

b) \(n_{H2SO4}=\dfrac{0,8.3}{2}=1,2\left(mol\right)\)

⇒ \(m_{H2SO4}=1,2.98=117,6\left(g\right)\)

\(m_{ddH2SO4}=\dfrac{117,6.100}{29,4}=400\left(g\right)\)

c) \(n_{Al2\left(SO4\right)3}=\dfrac{1,2.1}{3}=0,4\left(mol\right)\)

⇒ \(m_{Al2\left(SO4\right)3}=0,4.342=136,8\left(g\right)\)

\(m_{ddspu}=21,6+400-\left(1,2.2\right)=419,2\left(g\right)\)

\(C_{Al2\left(SO4\right)3}=\dfrac{136,8.100}{419,2}=32,63\)⁰/₀

 Chúc bạn học tốt

a) \(n_{Na_2CO_3}=\dfrac{21,2}{106}=0,2\left(mol\right)\)

PTHH: Na₂CO₃ + 2HCl --> 2NaCl + CO₂ + H₂O

                0,2------------------>0,4---->0,2

m_{dd sau pư} = 200 + 21,2 - 0,2.44 = 212,4(g)

=> \(C\%\left(NaCl\right)=\dfrac{0,4.58,5}{212,4}.100\%=11,017\%\)

$PTHH:Na_2CO_3+2HCl\to 2NaCl+H_2O+CO_2\uparrow$

$n_{Na_2CO_3}=\dfrac{21,2}{106}=0,2(mol)$

Theo PT: $n_{NaCl}=n_{CO_2}=0,2(mol)$

$\Rightarrow m_{NaCl}=0,4.58,5=23,4(g);m_{CO_2}=0,2.44=8,8(g)$

$\Rightarrow C\%_{NaCl}=\dfrac{23,4}{21,2+200-8,8}.100\%\approx 11,01\%$

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{H_2}=n_{ZnCl_2}=n_{Zn}=0,2\left(mol\right)\\ a,V_{H_2\left(Đktc\right)}=0,2.22,4=4,48\left(l\right)\\ b,m_{ZnCl_2}=136.0,2=27,2\left(g\right)\\ c,n_{HCl}=0,2.2=0,4\left(mol\right)\\ C\%_{ddHCl}=\dfrac{0,4.36,5}{200}.100\%=7,3\%\)

\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\ pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\) 
          0,1           0,2            0,1      0,1 
\(m_{HCl}=0,2.36,5=7,3\left(g\right)\\ V_{H_2}=0,1.22,4=2,24l\\ m_{\text{dd}}=6,5+200-\left(0,1.2\right)=206,3g\)  
bài 2 :
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ pthh:Mg+2HCl\rightarrow MgCl_2+H_2\) 
          0,2             0,4       0,2              0,2 
\(m_{HCl}=0,4.36,5=14,6g\\ V_{H_2}=0,2.22,4=4,48l\\ m\text{dd}=4,8+200-0,4=204,4g\\ C\%=\dfrac{0,2.136}{204,4}.100\%=13,3\%\)

\(a.n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\\ PTHH:Fe+2HCl\xrightarrow[]{}FeCl_2+H_2\)

tỉ lệ        :1        2            1            1

số mol   :0,1     0,2         0,1          0,1

\(V_{H_2}=0,1.22,4=2,24\left(l\right)\\ b)m_{HCl}=0,2.36,5=7,3\left(g\right)\\ C_{\%HCl}=\dfrac{7,3}{500}\cdot100\%=1,46\%\)

\(a,Fe+2HCl\rightarrow FeCl_2+H_2\\ CuO+2HCl\rightarrow CuCl_2+H_2O\\ n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\Rightarrow n_{Fe}=n_{H_2}=0,1\left(mol\right)\\ \Rightarrow\%m_{Fe}=\dfrac{0,1.56}{13,6}.100\%\approx41,176\%\\ \Rightarrow\%m_{CuO}\approx58,824\%\\ b,n_{CuO}=\dfrac{13,6-0,1.56}{80}=0,1\left(mol\right)\\ n_{HCl\left(p.ứ\right)}=2.\left(n_{Fe}+n_{CuO}\right)=2.\left(0,1+0,1\right)=0,4\left(mol\right)\\ \Rightarrow V_{ddHCl}=\dfrac{0,4}{2}=0,2\left(l\right)\)

\(n_{H_2}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)

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

\(0.1......0.2....................0.1\)

\(m_{Cu}=m_{hh}-m_{Fe}=12-0.1\cdot56=6.4\left(g\right)\)

\(C_{M_{HCl}}=\dfrac{0.2}{0.2}=1\left(M\right)\)

Bài 3 : 

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

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

        2            3                   1               3

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

a) \(n_{H2}=\dfrac{0,1.3}{2}=0,15\left(mol\right)\)

\(m_{H2}=0,15.2=0,3\left(g\right)\)

\(V_{H2\left(dktc\right)}=0,15.22,4=3,36\left(l\right)\)

b) \(n_{H2SO4}=\dfrac{0,1.3}{2}=0,15\left(mol\right)\)

⇒ \(m=0,15.98=14,7\left(g\right)\)

\(C_{ddH2SO4}=\dfrac{14,7.100}{200}=7,35\)⁰/₀

c) \(n_{Al2\left(SO4\right)3}=\dfrac{0,15.1}{3}=0,05\left(mol\right)\)

⇒ \(m_{Al2\left(SO4\right)3}=0,05.342=17,1\left(g\right)\)

\(m_{ddspu}=2,7+200-0,3=302,4\left(g\right)\)

\(C_{Al2\left(SO4\right)3}=\dfrac{17,1.100}{302,4}=5,65\)⁰/₀

 Chúc bạn học tốt

Fe+H2SO4->FeSO4+H2

0,1----0,1-------0,1-----0,1 

n Fe=5,6\56=0,1 mol

=>VH2=0,1.22,4=2,24l

=>m H2SO4=0,1.98=9,8g

=>C%=9,8\100 .100=9,8%