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

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

\(n_{K_2CO_3}=\dfrac{200.13,8\%}{100\%.138}=0,2(mol)\\ a,K_2CO_3+2HCl\to 2KCl+H_2O+CO_2\uparrow\\ b,n_{CO_2}=n_{K_2CO_3}=0,2(mol)\\ \Rightarrow V_{CO_2}=0,2.22,4=4,48(l)\\ c,n_{HCl}=0,4(mol)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{0,4.36,5}{14,6\%}=100(g)\)

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

rất mog nhận đc sự giúp đỡ từ mn

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

a) Fe + 2HCl --> FeCl₂ + H₂

b) \(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

         0,15-->0,3---->0,15-->0,15

=> V_H2 = 0,15.22,4 = 3,36 (l)

c) \(m_{dd.HCl}=\dfrac{0,3.36,5}{7,3\%}=150\left(g\right)\)

\(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\\ pthh:Fe+2HCl\rightarrow FeCl_2+H_2\)
           0,15    0,3                    0,15 
\(V_{H_2}=0,15.22,4=3,36l\\ m_{\text{dd}HCl}=\dfrac{100.\left(0,3.36,5\right)}{7,3}=150g\)

a)

$AgNO_3 + HCl \to AgCl + HNO_3$
Theo PTHH : 

$n_{AgCl} = n_{HCl} = n_{AgNO_3} = \dfrac{340.10\%}{170} =0,2(mol)$

$m_{dd\ HCl} = \dfrac{0,2.36,5}{7,3\%} = 100(gam)$

b)

$m_{AgCl} = 0,2.143,5 = 28,7(gam)$

c)

$m_{dd\ sau\ pư} = 340 + 100 -28,7 = 411,3(gam)$

$n_{HNO_3} = n_{AgNO_3} = 0,2(mol)$
$\Rightarrow C\%_{HNO_3} = \dfrac{0,2.63}{411,3}.100\% = 3,06\%$

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

PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

Theo PT: \(n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,2\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,2.24,79=4,958\left(l\right)\)

\(m_{ZnCl_2}=0,2.136=27,2\left(g\right)\)

b, \(n_{HCl}=2n_{Zn}=0,4\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,4.36,5=14,6\left(g\right)\Rightarrow m_{ddHCl}=\dfrac{14,6}{7,3\%}=200\left(g\right)\)