\(a,n_{K_2CO_3}=\dfrac{20,7}{138}=0,15(mol)\\ PTHH:K_2CO_3+2HCl\to 2KCl+H_2O+CO_2\uparrow\\ \Rightarrow n_{HCl}=2n_{K_2CO_3}=0,3(mol)\\ \Rightarrow m_{HCl}=0,3.36,5=10,95(g)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{10,95}{7,3\%}=150(g)\\ b,n_{CO_2}=n_{K_2CO_3}=0,15(mol)\\ \Rightarrow V_{CO_2}=0,15.22,4=3,36(l)\)

\(c,n_{KCl}=n_{HCl}=0,3(mol);n_{H_2O}=n_{CO_2}=0,15(mol)\\ \Rightarrow m_{KCl}=0,3.74,5=22,35(g)\\ m_{H_2O}=0,15.18=2,7(g);m_{CO_2}=0,15.44=6,6(g)\\ \Rightarrow m_{dd_{KCl}}=20,7+150-2,7-6,6=161,4(g)\\ \Rightarrow C\%_{KCl}=\dfrac{22,35}{161,4}.100\%\approx13,85\%\\ d,PTHH:CO_2+Ba(OH)_2\to BaCO_3\downarrow+H_2O\\ \Rightarrow n_{BaCO_3}=n_{CO_2}=0,15(mol)\\ \Rightarrow m_{BaCO_3}=0,15.197=29,55(g)\\ \Rightarrow m_{BaCO_3(\text{Thực tế})}=29,55.75\%=22,1625(g)\)

a) K₂CO₃ + 2HCl --> 2KCl + CO₂ + H₂O

b) \(n_{K_2CO_3}=\dfrac{13,8}{138}=0,1\left(mol\right)\)

PTHH: K₂CO₃ + 2HCl --> 2KCl + CO₂ + H₂O

______0,1----->0,2------>0,2--->0,1

=> V_CO2 = 0,1.22,4 = 2,24 (l)

c) m_HCl = 0,2.36,5 = 7,3 (g)

\(m_{ddHCl}=\dfrac{7,3.100}{7,3}=100\left(g\right)\)

d) m_KCl = 0,2.74,5 = 14,9 (g)

m_{dd sau pư} = 13,8 + 100 - 0,1.44 = 109,4 (g)

=> \(C\%\left(KCl\right)=\dfrac{14,9}{109,4}.100\%=13,62\%\)

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

PTHH: Na₂CO₃ + 2CH₃COOH --> 2CH₃COONa + CO₂ + H₂O

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

=> V_CO2 = 0,2.22,4 = 4,48 (l)

b) \(m_{dd.CH_3COOH}=\dfrac{0,4.60}{12\%}=200\left(g\right)\)

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

=> \(C\%_{muối}=\dfrac{0,4.82}{212,4}.100\%=15,44\%\)

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

a) Theo Pt : \(n_{H2}=n_{Fe}=n_{FeCl2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

\(\%m_{Fe}=\dfrac{0,1.56}{10}.100\%=56\%\)

\(\%m_{Cu}=100\%-56\%=44\%\)

b) Theo Pt : \(n_{H2}=2n_{HCl}=2.0,1=0,2\left(mol\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{0,2.36,5}{7,3\%}.100\%=100\left(g\right)\)

c) \(m_{ddspu}=10+100-0,1.2=109,8\left(g\right)\)

\(C\%_{FeCl2}=\dfrac{0,1.127}{109,8}.100\%=11,57\%\)

\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ a.Mg+2HCl\rightarrow MgCl_2+H_2\\ n_{H_2}=n_{MgCl_2}=n_{Mg}=0,2\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\\ b.n_{HCl}=0,2.2=0,4\left(mol\right)\\ m_{ddHCl}=\dfrac{0,4.36,5.100}{7,3}=200\left(g\right)\\ c.m_{ddsau}=4,8+200-0,2.2=204,4\left(g\right)\\ C\%_{ddMgCl_2}=\dfrac{0,2.95}{204,4}.100\approx9,295\%\\ d.V_{ddHCl}=\dfrac{200}{1,05}=\dfrac{4000}{21}\left(ml\right)=\dfrac{4}{21}\left(l\right)\\ C_{MddHCl}=\dfrac{0,4}{\dfrac{4}{21}}=2,1\left(M\right)\)

\(n_{Zn}=\dfrac{26}{65}=0,4\left(mol\right)\\ pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\) 
           0,4       0,8        0,4           0,4
\(a,V_{H_2}=0,4.22,4=8,96\left(l\right)\\ b,C\%_{HCl}=\dfrac{0,8.36,5}{150}.100\%=19,5\%\\ c,m_{\text{dd}}=26+150-\left(0,4.2\right)=175,2\left(g\right)\\ C\%_{ZnCl_2}=\dfrac{0,4.136}{175,2}.100\%=31\%\)

Tóm tắt

\(V_{H_2\left(đktc\right)}=8,96l\\ C_{\%H_2SO_4}=19,6\%\\ a)m_{Zn}=?\\ m_{ddH_2SO_4}=?\\ b)C_{\%ZnSO_4}=?\)

\(a)n_{H_2}=\dfrac{8,96}{22,4}=0,4mol\\ Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

0,4        0,4              0,4           0,4

\(m_{Zn}=0,4.65=26g\\ m_{ddH_2SO_4}=\dfrac{0,4.98}{19,6}\cdot100=200g\\ b)C_{\%ZnSO_4}=\dfrac{0,4.161}{26+200-0,4.2}\cdot100=28,6\%\)

\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

PTHH :

\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)

 0,4    0,4             0,4        0,4

\(a,m_{Zn}=0,4.65=26\left(g\right)\)

\(m_{H_2SO_4}=0,4.98=39,2\left(g\right)\)

\(m_{ddH_2SO_4}=\dfrac{39,2.100}{19,6}=200\left(g\right)\)

\(b,m_{ZnSO_4}=0,4.161=64,4\left(g\right)\)

\(m_{ddZnSO_4}=200+26-\left(0,4.2\right)=225,2\left(g\right)\)

\(C\%_{ZnSO_4}=\dfrac{64,4}{225,2}.100\%\simeq28,6\%\)

a)

$Mg + 2HCl \to MgCl_2 + H_2$

Theo PTHH : 

$n_{MgCl_2} = n_{H_2} = \dfrac{5,6}{22,4} = 0,25(mol)$
$m_{MgCl_2} = 0,25.95 = 23,75(gam)$

b)

$n_{HCl} = 2n_{H_2} = 0,5(mol)$
$m_{dd\ HCl} = \dfrac{0,5.36,5}{7,3\%} = 250(gam)$

c)

$2K + 2H_2O \to 2KOH + H_2$
$n_K = 2n_{H_2} = 0,5(mol)$
$m_K = 0,5.39 = 19,5(gam)$

Câu 1 :

a)  PTHH : 

 \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\) (1) 

  \(H_2+CuO\underrightarrow{t^o}Cu+H_2O\)(2)

b) Ta có : \(n_{Zn}=\frac{3,5}{65}\approx0,054\left(mol\right)\)

Theo phương trình hóa học (1) :

\(n_{H_2}=n_{Zn}\approx0,054\left(mol\right)\)

\(\Rightarrow V_{H_2}\approx0,054\cdot22,4=1,2096\left(l\right)\)

c) Theo phương trình hóa học (2), ta có:

\(n_{Cu}=n_{H_2}\approx0,054\left(mol\right)\)

\(\Rightarrow m_{Cu}\approx0,054\cdot64=3,456\left(g\right)\)

Bài 2:

a) Ta có : \(n_{Al}=\frac{2,7}{27}=0,1\left(mol\right)\); \(n_{HCl}=\frac{200\cdot7,3}{100\cdot36,5}=0,4\left(mol\right)\)

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

Theo phương trình hóa học : \(n_{H_2}=\frac{3}{2}n_{Al}=\frac{3}{2}.0,1=0,15\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,15\cdot22,4=3,36\left(l\right)\)

b) Theo phương trình hóa học , ta có : \(n_{AlCl_3}=n_{Al}=0,1\left(mol\right)\)

\(\Rightarrow m_{AlCl_3}=0,1\cdot133,5=13,35\left(g\right)\)

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

Áp dụng định luật bảo toàn khối lượng, ta có:

\(m_{Al}+m_{dd_{HCl}}=m_{dd_{AlCl_3}}+m_{H_2}\)

\(\Leftrightarrow2,7+200=m_{dd_{AlCl_3}}+0,3\)

\(\Leftrightarrow m_{dd_{AlCl_3}}=202,4\left(g\right)\)

Vậy \(C\%_{dd_{AlCl_3}}=\frac{13,35}{202,4}\cdot100\%\approx6,6\%\)