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nH2 = \(\frac{3,36}{22,4}\)= 0,15 mol
Fe + 2HCl➞ FeCl2 + H2
0,15 0,15
=> mFe = 0,15.56=8,4 gam
Cu không tác dụng với HCl
=> mCu = 10 - mFe = 1,6 gam
=> %mFe = \(\frac{8,4}{10}\) = 84 %
=> %mCu = 100 - 84 = 16 %
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{H2}=\frac{1}{2}n_{HCl}=0,25\left(mol\right)=>m_{H2}=0,5\left(g\right)\)
m dd sau pư =\(m_{KL}+m_{ddHCl}-m_{Crắn}-m_{H2}\)
\(=10,8+91,25-0,5-12,8=88,75\left(g\right)\)
\(C\%_{MgCl2}=\frac{0,1.95}{88,75}.100\%=10,7\%\)
\(C\%_{AlCl3}=\frac{0,15.133,5}{88,75}.100\%=22,56\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
15
a)\(Fe+H2SO4-->FeSO4+H2\)
\(n_{H2}=\frac{33,6}{22,4}=1,5\left(mol\right)\)
\(n_{Fe}=n_{H2}=1,5\left(mol\right)\)
\(m_{Fe}=1,5.56=84\left(g\right)\)
b)\(n_{FeSO4}=n_{H2}=1,5\left(mol\right)\)
\(m=m_{FeSO4}=1,5.152=228\left(g\right)\)
c)\(n_{H2SO4}=n_{H2}=1,5\left(mol\right)\)
\(C_{M\left(H2SO4\right)}=\frac{1,5}{0,5}=3\left(M\right)\)
16.
n\(_{H2}=\frac{2,24}{22,4}=0,1\left(mol\right)\)
Gọi \(n_{Mg}=x,n_{Fe}=y\)
\(Mg+2HCl--.MgCl2+H2\)
x-------------------------x----------x(mol)
\(Fe=2HCl-->FeCl2+H2\)
y----------------------------y------y(mol)
Theo bài ta có hpt
\(\left\{{}\begin{matrix}24x+56y=4\\x+y=0,1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,05\\y=0,05\end{matrix}\right.\)
\(m_{MgCl2}=0,05.95=4,75\left(g\right)\)
\(m_{FeCl2}=0,05.127=6,35\left(g\right)\)
17.
\(n_{H2}=\frac{13,44}{22,4}=0,6\left(mol\right)\)
\(Fe+2HCl--.FeCl2+H2\)
x----------------------------------x(mol)
\(2Al+6HCl--.2AlCl3+3H2\)
y----------------------------------------1,5y(mol)
theo bài ta có hpt
\(\left\{{}\begin{matrix}56x+27y=22,2\\x+1,5y=0,6\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,3\\y=0,2\end{matrix}\right.\)
\(\%m_{Fe}=\frac{0,3.56}{22,2}.100\%=75,68\%\%\)
\(\%m_{Al}=100-75,68=24,32\%\)
18.
\(Mg+2HCl--.MgCl2+H2\)
\(Fe+2HCl--.FeCl2+H2\)
Chất rắn k tan là Cu = 2,54(g)
=>\(m_{Mg+Fe}=10,54-02,54=10\left(g\right)\)
\(n_{H2}=\frac{4,48}{22,4}=0,2\left(mol\right)\Rightarrow m_{H2}=0,4\left(g\right)\)
\(n_{HCl}=n_{H2}=0,4\left(mol\right)\Rightarrow m_{HCl}=0,4.36,5=14,6\left(g\right)\)
\(m=m_{Fe+Mg}+m_{HCl}-m_{H2}=10+14,6-0,4=24,2\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : nH2 = \(\dfrac{0,2}{2}\)= 0,1 ( mol )
Mg + 2HCl → MgCl2 + H2
x → 2x → x → x
Fe + 2HCl → FeCl2 + H2
y → 2y → y → y
Còn Cu không tác dụng được với HCl => 0,64 gam chất rắn là Cu
=> mMg + mFe = 4,64 - 0,64 = 4 ( gam )
=> \(\left\{{}\begin{matrix}24x+56y=4\\x+y=0,1\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=0,05\\y=0,05\end{matrix}\right.\)
=> mFe = 56 . 0,05 = 2,8 ( gam )
=> mMg = 24 . 0,05 = 1,2 ( gam )
=> mMgCl2 = 95 . 0,05 = 4,75 ( gam )
=> mFeCl2 = 127 . 0,05 = 6,35 ( gam )
![](https://rs.olm.vn/images/avt/0.png?1311)
nH2 = \(\dfrac{5,6}{22,4}\)= 0,25 (mol)
a, Mg + 2HCl ----> MgCl2 + H2
0,25 0,5 0,25 0,25 (mol)
=> mMg = 0,25.24 = 6 (g)
=> mCu = 16 - 6 = 10 (g)
%Mg = \(\dfrac{6.100\%}{16}\)= 37,5%
%Cu = 100 - 37,5 = 62,5%
b,
mHCl = 0,5.36,5 = 18,25 (g)
=> mdd HCl = \(\dfrac{18,25.100\%}{7,3\%}\)= 250 (g)
c,
Mg + CuSO4 ----> MgSO4 + Cu
0,25 0,25 0,25 0,25 (mol)
=> mCu = 0,25.64 + 10 = 26 (g)
=> mCuSO4 = 0,25.160 = 40 (g)
=> mdd CuSO4 = \(\dfrac{40.100\%}{20\%}\)= 200 (g)
![](https://rs.olm.vn/images/avt/0.png?1311)
Fe + 2HCl -> FeCl2 + H2 (1)
nHCl=0,4.1=0,4(mol)
%mCu=\(\dfrac{6,4}{12}.100\%=53,3\%\)
%mFe=100-53,3=46,7%
mFe=12-6,4=5,6(g)
nFe=0,1(mol)
b;KOH + HCl -> KCl + H2O (2)
2KOH + FeCl2 -> Fe(OH)2 + 2KCl (3)
nKOH=0,36.1=0,36(mol)
nHCl (1)=2nFe=0,2(mol)
nHCl còn lại=0,4-0,2=0,2(mol)<0,36 nên có PƯ 3
Theo PTHH 2 ta có:
nKOH tham gia PƯ=nHCl=0,2(mol)
=>nKOH còn lại=0,36-0,2=0,16(mol)
VÌ 2nFeCl2=0,2>nKOH=0,16 nên FeCl2 dư
Theo PTHH 3 ta có:
\(\dfrac{1}{2}\)nKOH=nFe(OH)2=0,08(mol)
mFe(OH)2=0,08.90=7,2(g)