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1.1. Al + NaOH + H2O ==> NaAlO2 + 3/2H2
nH2(1)=3,36/22,4=0.15(mol)
=> nAl(1)= nH2(1):3/2= 0.15:3/2= 0.1(mol)
2.Mg + 2HCl ==> MgCl2 + H2
3.2Al + 6HCl ==> 2AlCl3 + 3H2
4.Fe + 2HCl ==> FeCl2 + H2
=> \(n_{H_2\left(2,3,4\right)}=\) 10.08/22.4= 0.45(mol)
=> nH2(3)=0.1*3/2=0.15(mol)
MgCl2 + 2NaOH ==> Mg(OH)2 + 2NaCl
AlCl3 + 3NaOH ==> Al(OH)3 + 3NaCl
FeCl2 + 2NaOH ==> Fe(OH)2 + 2NaCl
a, Giả sử: \(\left\{{}\begin{matrix}n_{CO_2}=x\left(mol\right)\\n_{SO_2}=y\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow x+y=\dfrac{0,224}{22,4}=0,01\left(1\right)\)
Mà: \(\overline{M}_A=56\Rightarrow44x+64y=56.0,01\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,004\left(mol\right)\\y=0,006\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%n_{CO_2}=\dfrac{0,004}{0,01}.100\%=40\%\\\%n_{SO_2}=60\%\end{matrix}\right.\)
BTNT C và S, có: \(\left\{{}\begin{matrix}n_{Na_2CO_3}=n_{CO_2}=0,004\left(mol\right)\\n_{Na_2SO_3}=n_{SO_2}=0,006\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Na_2CO_3}=\dfrac{0,004.106}{0,004.106+0,006.126}.100\%\approx35,9\%\\\%m_{Na_2SO_3}\approx64,1\%\end{matrix}\right.\)
b, Ta có: \(n_{HCl}=0,05.0,2=0,01\left(mol\right)\)
PT: \(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)
____0,005_______0,01 (mol)
\(Ba\left(OH\right)_2+CO_2\rightarrow BaCO_3+H_2O\)
_0,004______0,004 (mol)
\(Ba\left(OH\right)_2+SO_2\rightarrow BaSO_3+H_2O\)
_0,006_____0,006 (mol)
\(\Rightarrow n_{Ba\left(OH\right)_2}=0,015\left(mol\right)\)
\(\Rightarrow C_{M_{Ba\left(OH\right)_2}}=\dfrac{0,015}{1}=0,015M\)
Bạn tham khảo nhé!
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)\\ Mg+2HCl\to MgCl_2+H_2\\ MgO+2HCl\to MgCl_2+H_2O\\ \Rightarrow n_{Mg}=0,1(mol)\\ \Rightarrow \%_{Mg}=\dfrac{0,1.24}{6,4}.100\%=37,5\%\\ \Rightarrow \%_{MgO}=100\%-37,5\%=62,5\%\)
\(b,n_{MgO}=\dfrac{6,4-0,1.24}{40}=0,1(mol)\\ \Rightarrow n_{HCl}=2.0,1+2.0,1=0,4(mol)\\ \Rightarrow V_{dd_{HCl}}=\dfrac{0,4}{0,5}=0,8(l)\\ c,n_{MgCl_2}=0,1+0,1=0,2(mol)\\ \Rightarrow C_{M_{MgCl_2}}=\dfrac{0,2}{0,8}=0,25M\)
nH2 = \(\frac{4,48}{22,4}\)= 0,2 mol
PTHH:
Fe + 2HCl\(\rightarrow\) FeCl2 + H2
FeO + 2HCl \(\rightarrow\) FeCl2 + H2O
\(\rightarrow\) nFe = nH2 = 0,2
\(\rightarrow\)mFe = 0,2.56=11,2 g \(\rightarrow\)mFeO = 18,3 -11,2 = 7,2 g
\(\rightarrow\) nFeO =\(\frac{7,2}{72}\) = 0,1 mol
nHCl = 2 (nFe+nFeO) = 0,6 mol
\(\Rightarrow\) mHCl = 36,5 .0,6 = 21,9
\(\Rightarrow\) C%HCl = \(\frac{21,9}{200}.100\%\) = 43,8%
Bảo toàn khối lượng :
mddsaupứ = mFe + mFeO + mddHCl - mH2
= 18,4 + 200 - 0,4 = 218 g
nFeCl2 = nFe + nFeO = 0,3 mol
mFeCl2 = 127. 0,3 = 38,1 g
C%FeCl2 = \(\frac{38,1}{218}.100\%\) = 17,48%
Theo de bai ta co : nH2 = \(\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
Ta co PTHH :
(1) Fe+ 2HCl \(->FeCl2+H2\uparrow\)
0,1 mol.....................................0,1mol
(2) \(Fe2O3+6HCl->2FeCl3+3H2O\)
a) ta cos :
mFe = 0,1.56 = 5,6 (g)
=> %mFe = \(\dfrac{5,6}{28,8}.100\%\approx19,44\%\)
%mFe2O3 = 100% - 19,44% = 80,56%
b) Theo PTHH 1 va 2 ta co :
nHCl = 2nH2 = 0,2 (mol)
Ta co PTHH :
| 16HCl | + | 2KMnO4 | → | 5Cl2 | + | 8H2O | + | 2KCl | + | 2MnCl2 |
| 0,2mol | 0,025(mol) | |||||||||
=> VddKMnO4 = \(\dfrac{0,025}{1}=0,025\left(l\right)\)
Ta có nH2 = \(\dfrac{2,24}{22,4}\) = 0,1 ( mol )
Fe + 2HCl \(\rightarrow\) FeCl2 + H2
x.........2x...........x...........x
Fe3O4 + 8HCl \(\)\(\rightarrow\) FeCl2 + 2FeCl3 + 4H2
y................8y..........y..............2y..........4y
=> \(\left\{{}\begin{matrix}56x+232y=28,8\\x+4y=0,1\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=-11,5\\y=2,9\end{matrix}\right.\)
Hình như đề sai bạn ơi