\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\) 
          0,2        0,4                      0,2 
\(V_{H_2}=0,2.22,4=4,48\left(l\right)\\ C\%_{HCl}=\dfrac{0,4.36,5}{200}.100\%=7,3\%\)

Fe+H2SO4->FeSO4+H2

0,15---0,15-----0,15---0,15 mol

n Fe=8,4\56=0,15 mol

=>VH2=0,15.22,4=3,36l

=>m H2SO4=0,15.98=14,7g

=>C% H2SO4=14,7\245 .100=6%

=>m dd muối=8,4+245-0,15.2=253,1g

=>C% muối =0,15.152\253,1 .100=9%

\(n_{AlCl_3}=\dfrac{26.7}{133.5}=0.2\left(mol\right)\)

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

\(.........0.6......0.2.......0.3\)

\(V_{H_2}=0.3\cdot22.4=6.72\left(l\right)\)

\(C\%HCl=\dfrac{0.6\cdot36.5}{300}\cdot100\%=7.3\%\)

\(n_{Fe}=\dfrac{84}{56}=1,5\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{H_2}=n_{FeCl_2}=n_{Fe}=1,5\left(mol\right)\\ V_{H_2}=1,5.22,4=33,6\left(l\right)\\ C\%_{ddFeCl_2}=\dfrac{127.1,5}{84+300-1,5.2}.100\%=\dfrac{190,5}{381}.100\%=50\%\)

\(n_{HCl}=0,3.2=0,6\left(mol\right)\\a, 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ b,n_{H_2}=\dfrac{3}{6}.0,6=0,3\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\\ c,n_{Al}=n_{AlCl_3}=\dfrac{2}{6}.0,6=0,2\left(mol\right)\\ m_{Al}=0,2.27=5,4\left(g\right)\\ d,V_{ddAlCl_3}=V_{ddHCl}=0,3\left(l\right)\\ C_{MddHCl}=\dfrac{0,2}{0,3}=\dfrac{2}{3}\left(M\right)\)

Bài 4 : 

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

a) Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2|\)

            1          2              1          1

          0,2        0,4           0,2        0,2

b) \(n_{H2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

\(V_{H2\left(dktc\right)}=0,2.22,4=4,48\left(l\right)\)

c) \(n_{HCl}=\dfrac{0,2.2}{1}=0,4\left(mol\right)\)

⇒ \(m_{HCl}=0,4.36,5=14,6\left(g\right)\)

\(C_{ddHCl}=\dfrac{14,6.100}{100}=14,6\)⁰/₀

d) \(n_{ZnCl2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

⇒ \(m_{ZnCl2}=0,2.136=27,2\left(g\right)\)

\(m_{ddspu}=13+100-\left(0,2.2\right)=112,6\left(g\right)\)

\(C_{ZnCl2}=\dfrac{27,2.100}{112,6}=24,16\)⁰/₀

 Chúc bạn học tốt

Bài 3 : 

\(n_{Mg}=\dfrac{12}{24}=0,5\left(mol\right)\)

a) Pt : \(Mg+H_2SO_4\rightarrow MgSO_4+H_2|\)

            1            1                1           1

          0,5          0,5            0,5          0,5

b) \(n_{H2}=\dfrac{0,5.1}{1}=0,5\left(mol\right)\)

\(V_{H2\left(dktc\right)}=0,5.22,4=11,2\left(l\right)\)

c) \(n_{H2SO4}=\dfrac{0,5.1}{1}=0,5\left(mol\right)\)

⇒ \(m_{H2SO4}=0,5.98=49\left(g\right)\)

\(C_{ddH2SO4}=\dfrac{49.100}{200}=24,5\)⁰/₀

d) \(n_{MgSO4}=\dfrac{0,5.1}{1}=0,5\left(mol\right)\)

⇒ \(m_{MgSO4}=0,5.120=60\left(g\right)\)

\(m_{ddspu}=12+200-\left(0,5.2\right)=211\left(g\right)\)

\(C_{MgSO4}=\dfrac{60.100}{211}=28,44\)⁰/₀

 Chúc bạn học tốt

Mg+H2SO4->MgSO4+H2

0,5----0,5--------0,5--------0,5

n Mg=12\24=0,5 mol

=>VH2=0,5.22,4=11,2l

=>m H2SO4=0,5.98=49 g

=>C% =49\200.100=24,5g

=>m dd=12+200-0,5.2=211g

=>C%=0,4.120\211.100=22,74885%

\(a,n_{Al_2O_3}=\dfrac{20,4}{102}=0,2\left(mol\right)\\ PTHH:Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\\ .......0,2......1,2......0,4.....0,6\left(mol\right)\\ V_{H_2O\left(đktc\right)}=0,6\cdot22,4=13,44\left(l\right)\\ b,m_{HCl}=36,5\cdot1,2=43,8\left(g\right)\\ c,m_{dd_{HCl}}=1,2\cdot36,5=43,8\left(g\right)\\ C\%_{dd_{HCl}}=\dfrac{43,8}{200}\cdot100\%=21,9\%\)

\(d,m_{AlCl_3}=0,4\cdot133,5=53,4\left(g\right)\\ m_{dd\left(sau.pứ\right)}=m_{Al_2O_3}+m_{dd_{HCl}}-m_{H_2}=20,4+200-10,8=209,6\left(g\right)\\ C\%_{dd_{AlCl_3}}=\dfrac{53,4}{209,6}\cdot100\%\approx25,48\%\)

d)\(m_{NaOH}=\dfrac{mct.C\%}{100}=\dfrac{100.20}{100}=20\left(g\right)\)=>\(n_{NaOH}=\dfrac{20}{40}=0,5\left(mol\right)\)

\(HCl+NaOH\rightarrow NaCl+H_2O\)

tl1............1.............1..............1(mol)

So sánh\(\dfrac{n_{NaOH}}{1}< \dfrac{n_{HCl}}{1}\left(\dfrac{0,5}{1}< \dfrac{1,2}{1}\right)\)

=> HCl Dư

Vậy Dung dịch sau phản ứng làm quỳ tím chuyển màu đỏ

a)

$n_{Al} = \dfrac{8,1}{27} = 0,3(mol)$

$2Al + 6HCl \to 2AlCl_3 + 3H_2$
Theo PTHH : 

$n_{H_2} = \dfrac{3}{2}n_{Al} = 0,45(mol)$
$V_{H_2} = 0,45.22,4 = 10,08(lít)$

b) $n_{HCl} = 3n_{Al} = 0,9(mol)$
$\Rightarrow m_{dd\ HCl} = \dfrac{0,9.36,5}{3,65\%} = 900(gam)$

c)

$m_{dd\ sau\ pư}= 8,1 + 900 - 0,45.2 = 907,2(gam)$

$n_{Al_2(SO_4)_3} = \dfrac{1}{2}n_{Al} = 0,15(mol)$
$C\%_{Al_2(SO_4)_3} = \dfrac{0,15.342}{907,2}.100\% = 5,65\%$

Fe + 2HCl -> FeCl2 + H2
nFe = 5,6/56 = 0,1 mol
=>nH2 = 0,1 mol
=> VH2= 0,1*22,4= 2,24 lít

\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

PTHH: Fe + 2HCl ---> FeCl₂ + H₂

          0,1-->0,2------------------>0,1

=> \(\left\{{}\begin{matrix}m_{ddHCl}=\dfrac{0,2.36,5}{15\%}=\dfrac{146}{3}\left(g\right)\\V_{H_2}=0,1.22,4=4,48\left(l\right)\end{matrix}\right.\)