Câu 1
\(a)PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\\ b)200ml=0,2l\\ n_{HCl}=0,2.1=0,2mol\\ n_{H_2}=n_{MgCl_2}=\dfrac{1}{2}n_{H_2}=\dfrac{1}{2}\cdot0,2=0,1mol\\ V_{H_2}=0,1.24,79=2,479l\\ c)C_{M_{MgCl_2}}=\dfrac{0,1}{0,2}=0,5M\)

Câu 2
\(a)PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\\ b)n_{Mg}=\dfrac{1,2}{24}=0,05mol\\ n_{H_2}=n_{MgCl_2}=n_{Mg}=0,05mol\\ V_{H_2}=0,05.24,79=1,2395l\\ c)n_{HCl}=2n_{Mg}=2.0,05=0,1mol\\ V_{ddHCl}=\dfrac{0,1}{2}=0,05l\\ d)C_{M_{MgCl_2}}=\dfrac{0,05}{0,05}=1M\)

\(n_{BaCl_2}=\dfrac{200.20,8\%}{208}=0,2\left(mol\right)\\ PTHH:BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl\\ n_{BaSO_4}=n_{H_2SO_4}=n_{BaCl_2}=0,2\left(mol\right)\\ a,m_{kt}=m_{BaSO_4}=233.0,2=46,6\left(g\right)\\ b,C\%_{ddH_2SO_4}=\dfrac{0,2.98}{200}.100\%=9,8\%\)

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

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

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

\(\Rightarrow C_MHCl=\dfrac{0,2}{0,5}=0,4M\)

Fe + 2HCl → FeCl₂ + H₂

n_Fe = \(\dfrac{5,6}{56}\) = 0,1 (mol)

Theo PT:n_HCl = 2n_Fe = 0,2 (mol)

⇒ C_MHCl = \(\dfrac{0,2}{0,5}\) = 0,4M

\(200ml=0,2l\\ n_{Na_2SO_4}=0,2.0,1=0,02mol\\ a)Na_2SO_4+BaCl_2\rightarrow BaSO_4+2NaCl\)
0,02                 0,02          0,02             0,04
\(b)m=m_{BaSO_4}=0,02.233=4,66g\\ c)50ml=0,05l\\ C_{M_{BaCl_2}}=\dfrac{0,02}{0,05}=0,4M\)

\(\left(1\right)C+O_2\xrightarrow[]{t^0}CO_2\\ \left(3\right)CO_2+Na_2O\rightarrow Na_2CO_3\\ \left(3\right)Na_2CO_3+CaCl_2\rightarrow CaCO_3+2NaCl\)

\(a.Fe+2HCl\rightarrow FeCl_2+H_2\\ b.H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\)

nH2 = 1,456:22,4 = 0,065

n Cl- = n HCl = 2nH2 = 0,13

m= m kloai + m cl- = 6,225 gam chọn B

\(n_{H_2}=\dfrac{1,456}{22,4}=0,065mol\\ BTNT\left(H\right):n_{HCl}=2n_{H_2}\\ \Leftrightarrow n_{HCl}=2.0,065=0,13mol\\ BTKL:m_{kl}+m_{HCl}=m_{muối}+m_{khí}\\ \Leftrightarrow1,61+0,13.36,5=m_{muối}+0,065.44\\ \Leftrightarrow m_{muối}=3,495g\)