\(Fe\left(NO_3\right)_3+3KOH\rightarrow Fe\left(OH\right)_3+3KNO_3\\ 2Fe\left(OH\right)_3\rightarrow\left(t^o\right)Fe_2O_3+3H_2O\\ Fe_2O_3+3CO\rightarrow\left(t^o\right)2Fe+3CO_2\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ FeCl_2+2KOH\rightarrow Fe\left(OH\right)_2+2KCl\)

Sửa đề chất cuối thành Fe(OH)₂

\(Muối:ACO_3,B_2CO_3\\ ACO_3+2HCl\rightarrow ACl_2+CO_2+H_2O\\ B_2CO_3+2HCl\rightarrow2BCl+CO_2+H_2O\\ n_{CO_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\Rightarrow m_{CO_2}=44.0,03=1,32\left(g\right)\\ n_{CO_3^{2-}}=n_{CO_2}=0,03\left(mol\right);n_{Cl^-}=2.n_{CO_2}=0,03.2=0,06\left(mol\right)\\ m_{muối.ddX}=m_{muối.cacbonat}+m_{Cl^-}-m_{CO_3^{2-}}=10+0,06.35,5-60.0,03=10,3\left(g\right)\)

\(a,Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{ZnCl_2}=n_{H_2}=n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\ b,m_{ZnCl_2}=136.0,1=13,6\left(g\right);V_{H_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\\ c,Vì:\dfrac{0,1}{1}>\dfrac{0,2}{1}\Rightarrow n_{Zn\left(TT\right)}=0,1\left(mol\right);n_{Zn\left(LT\right)}=n_{ZnCl_2}=0,2\left(mol\right)\\ H=\dfrac{0,1}{0,2}.100\%=50\%\)

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right);n_{ZnCl_2}=\dfrac{20,6}{136}\approx0,15147\left(mol\right)\\ PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ Vì:\dfrac{0,2}{1}>\dfrac{0,15147}{1}\\ Có:m_{ZnCl_2\left(LT\right)}=0,2.136=27,2\left(g\right)\\ H=\dfrac{20,6}{27,2}.100\%\approx75,735\%\)

a)

\(PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)

b)

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

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

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

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

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

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

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

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

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

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

(x²-2x+1)-(4y²-4y+1)

(x-1)²- (2y-1)²

(x-2y)(x+2y-2)

tích đúng cho mình nhé bn ơi

\(a)n_{H_2}=\dfrac{2,24}{22,4}=0,1mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\)
0,1       0,2          0,1           0,1
\(n_{Fe_2O_3}=\dfrac{21,6-56.0,1}{160}=0,1mol\\ Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2\)
0,1            0,6            0,2              0,3
\(V_{ddHCl}=\dfrac{0,2+0,6}{1}=0,8l\\ b.C_{M_{FeCl_2}}=\dfrac{0,1}{0,8}=0,125M\\ C_{M_{FeCl_3}}=\dfrac{0,2}{0,8}=0,25M\)