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![](https://rs.olm.vn/images/avt/0.png?1311)
\(a.CaO+2HCl\rightarrow CaCl_2+H_2O\\ n_{CaO}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ n_{CaCl_2}=n_{CaO}=0,2\left(mol\right)\\ m_{CaCl_2}=111.0,2=22,2\left(g\right)\)
Tên muối: Canxi clorua
\(b.n_{HCl}=0,2.2=0,4\left(mol\right)\\ V_{ddHCl}=\dfrac{0,4}{2}=0,2\left(l\right)=200\left(ml\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Fe}=\dfrac{2,8}{56}=0,05\left(mol\right)\)
PTHH :
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,05 0,1 0,05 0,05
\(V_{H_2}=0,05.22,4=1,12\left(l\right)\)
\(V_{HCl}=\dfrac{n}{C_M}=\dfrac{0,1}{1}=0,1\left(l\right)\)
\(m_{FeCl_2}=0,05.127=6,35\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Gọi số mol Mg, Ca là a, b
=> 24a + 40b = 8,8
PTHH: Mg + 2HCl --> MgCl2 + H2
______a---->2a------>a------->a
Ca + 2HCl --> CaCl2 + H2
b---->2b------->b------->b
=> a + b = \(\dfrac{6,72}{22,4}=0,3\)
=> a = 0,2 ; b = 0,1
=> \(\left\{{}\begin{matrix}\%Mg=\dfrac{0,2.24}{8,8}.100\%=54,55\%\\\%Ca=\dfrac{0,1.40}{8,8}.100\%=45,45\%\end{matrix}\right.\)
b) nHCl = 2a + 2b = 0,6 (mol)
=> \(V_{ddHCl}=\dfrac{0,6}{2}=0,3\left(l\right)\)
c) mMgCl2 = 0,2.95 = 19 (g)
mCaCl2 = 0,1.111 = 11,1 (g)
=> Tổng khối lượng muối = 19 + 11,1 = 30,1(g)
![](https://rs.olm.vn/images/avt/0.png?1311)
Fe+2HCl->FeCl2+H2
x---2x-----------x
Mg+2HCl->MgCl2+H2
y------2y-----------y
Ta có :
\(\left\{{}\begin{matrix}56x+24y=24\\x+y=\dfrac{13,44}{22,4}\end{matrix}\right.\)
=>x=0,3 mol, y=0,3 mol
=>%m Fe=\(\dfrac{0,3.56}{24}.100\)=70%
=>%m Mg=100-70=30%
=>VHCl=\(\dfrac{0,3.2+0,3.2}{2}\)=0,6l=600ml
b)
XCl2+2AgNO3->2AgCl+X(NO3)2
0,6--------------------1,2mol
=>m AgCl=1,2.143,5=172,2g
\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2|\)
1 2 1 1
0,1 0,2 0,1
a) \(n_{HCl}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)
\(m_{HCl}=0,2.36,5=7,3\left(g\right)\)
\(m_{ddHCl}=\dfrac{7,3.100}{7,3}=100\left(g\right)\)
b) \(n_{ZnCl2}=\dfrac{0,2.1}{2}=0,1\left(mol\right)\)
⇒ \(m_{ZnCl2}=0,1.136=13,6\left(g\right)\)
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