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

a) PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)

b) Ta có: \(n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right)=n_{MgCl_2}\)

\(\Rightarrow m_{MgCl_2}=0,15\cdot95=14,25\left(g\right)\)

c) Theo PTHH: \(\left\{{}\begin{matrix}n_{HCl\left(p.ứ\right)}=0,3\left(mol\right)\\n_{H_2}=0,15\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{HCl\left(p.ứ\right)}=0,3\cdot36,5=10,95\left(g\right)\\m_{H_2}=0,15\cdot2=0,3\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd}=m_{Mg}+m_{ddHCl}-m_{H_2}=53,3\left(g\right)\)

\(\Rightarrow m_{ddHCl}=50\left(g\right)\) \(\Rightarrow C\%_{HCl\left(p.ứ\right)}=\dfrac{10,95}{50}\cdot100\%=21,9\%\)  

nMg = 3,6 : 24 = 0,15 (mol) 
pthh : Mg + 2HCl --> MgCl2 + H2 
           0,15-----------> 0,15 --->0,15 (mol) 
mMgCl2 = 0,15 . 95 = 14,25 (mol) 
VH2 (đkc)= 0,15. 24,79 = 3,718(l) 
 

\(n_{Mg}=\dfrac{3,6}{24}=0,15mol\)

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

0,15      0,3         0,15       0,15

\(m_{MgCl_2}=0,15\cdot95=14,25g\)

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

\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ a,PTHH:Mg+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+H_2\\ n_{H_2}=n_{\left(CH_3COO\right)_2Mg}=n_{Mg}=0,2\left(mol\right);n_{CH_3COOH}=2.0,2=0,4\left(mol\right)\\ b,V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\\ c,m_{\left(CH_3COO\right)_2Mg}=142.0,2=28,4\left(g\right)\\ d,m_{ddCH_3COOH}=\dfrac{0,4.60.100}{12}=200\left(g\right)\)

\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\ pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\) 
           0,1     0,2                       0,1 
\(V_{H_2}=0,1.22,4=2,24\left(L\right)\\ V_{HCl}=\dfrac{0,2}{1}=0,2\left(L\right)\)

Zn+2HCl->ZnCl2+H2

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

n Zn=0,1 mol

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

=>VHCl=\(\dfrac{0,1}{1}=0,1l=100ml\)

a, PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

b, Ta có: \(n_{Fe}=\dfrac{14}{56}=0,25\left(mol\right)\)

Theo PT: \(n_{HCl}=2n_{Fe}=0,5\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,25.36,5=18,25\left(g\right)\)

c, Theo PT: \(n_{H_2}=n_{Fe}=0,25\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,25.22,4=5,6\left(l\right)\)

a) PTHH: Fe + 2HCl --> FeCl₂ + H₂

b) \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

            0,2-->0,4------>0,2-->0,2

=> \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)

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

d) \(m_{FeCl_2}=0,2.127=25,4\left(g\right)\)

\(a.n_{Mg\left(OH\right)_2}=\dfrac{17,4}{58}=0,3\left(mol\right)\\ Mg\left(OH\right)_2+2HCl\rightarrow MgCl_2+2H_2O\\ n_{HCl}=2n_{Mg\left(OH\right)_2}=0,6\left(mol\right)\\ CM_{HCl}=\dfrac{0,6}{0,2}=3M\\b. n_{Mg\left(OH\right)_2}=n_{MgCl_2}=0,3\left(mol\right)\\ m_{MgCl_2}=0,3.85=25,5\left(g\right)\\c.CM_{MgCl_2}=\dfrac{0,3}{0,2}=1,5M \)

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

b, \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

Theo PT: \(n_{H_2}=n_{Zn}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)

c, \(n_{HCl}=2n_{Zn}=0,2\left(mol\right)\Rightarrow m_{HCl}=0,2.36,5=7,3\left(g\right)\)

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

\(1mol\)   \(2mol\)                    \(1mol\)

\(0,1mol\)  \(0,2mol\)              \(0,1mol\)

\(n_{Zn}=\dfrac{m}{M}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

\(m_{HCl}=n.M=0,2.36,5=7,3\left(g\right)\)

\(V_{H_2}=n.22,4=0,1.22,4=2,24\left(l\right)\)