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\(n_{SO_2}=\dfrac{11,2}{22,4}=0,5mol\\ n_{Ca\left(OH\right)_2}=0,2.0,5=0,1mol\\ SO_2+Ca\left(OH\right)_2\rightarrow CaSO_3+H_2O\\ \Rightarrow\dfrac{0,5}{1}>\dfrac{0,1}{1}\Rightarrow SO_2.dư\\ n_{SO_2.pứ}=n_{Ca\left(OH\right)_2}=0,1mol\\ m_{SO_2.dư}=\left(0,5-0,1\right).64=25,6g\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
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
nFe = \(\dfrac{5,6}{56}\) = 0,1 (mol)
Theo PT:nHCl = 2nFe = 0,2 (mol)
⇒ CMHCl = \(\dfrac{0,2}{0,5}\) = 0,4M
\(Fe\left(NO_3\right)_3+3KOH\rightarrow Fe\left(OH\right)_3+3KNO_3\\ 2Fe\left(OH\right)_3\rightarrow\left(t^o\right)Fe_2O_3+3H_2O\\ Fe_2O_3+3CO\rightarrow\left(t^o\right)2Fe+3CO_2\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ FeCl_2+2KOH\rightarrow Fe\left(OH\right)_2+2KCl\)
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: \(n_{H_2}=n_{Zn}=0,2\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,2.24,79=4,958\left(l\right)\)
\(n_{BaCl_2}=\dfrac{200.20,8\%}{208}=0,2\left(mol\right)\\ PTHH:BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl\\ n_{BaSO_4}=n_{H_2SO_4}=n_{BaCl_2}=0,2\left(mol\right)\\ a,m_{kt}=m_{BaSO_4}=233.0,2=46,6\left(g\right)\\ b,C\%_{ddH_2SO_4}=\dfrac{0,2.98}{200}.100\%=9,8\%\)