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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
a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
Gọi \(\left\{{}\begin{matrix}n_{Fe}:a\left(mol\right)\\n_{Zn}:b\left(mol\right)\end{matrix}\right.\)
Giải hệ PT:
\(\left\{{}\begin{matrix}56a+65b=18,6\\127a+136b=39,9\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a=0,1\\b=0,2\end{matrix}\right.\)
\(\%m_{Fe}=\frac{0,1.56}{18,6}.100\%=30,11\%\)
\(\%m_{Fe}=100\%-30,11\%=69,89\%\)
b, \(n_{HCl}=0,1.2+0,2.2=0,6\left(mol\right)\)
\(V_{dd\left(HCl\right)}=\frac{0,6}{4}=0,15\left(l\right)=150\left(ml\right)\)
\(m_{dd\left(HCl\right)}=150.1,1=165\left(g\right)\)
c,\(ZnCl_2+2AgNO_3\rightarrow Zn\left(NO_3\right)_2+2AgCl\)
\(FeCl_2+2AgNO_3\rightarrow Fe\left(NO_3\right)_3+2AgCl\)
\(n_{AgCl}=0,1.2+0,2.2=0,6\left(mol\right)\)
\(\Rightarrow m_{AgCl}=0,6.143,5=86,1\left(g\right)\)
a, \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
PTHH: Zn + 2HCl → ZnCl2 + H2
Mol: 0,1 0,1 0,1
PTHH: CuO + 2HCl → CuCl2 + H2O
Mol: 0,01 0,01
\(\%m_{Zn}=\dfrac{0,1.65.100\%}{7,3}=89,04\%\)
\(\%m_{CuO}=100-89,04=10,96\%\)
b, \(n_{CuO}=\dfrac{7,3-0,1.65}{80}=\dfrac{0,8}{80}=0,01\left(mol\right)\)
\(m_{muối}=0,1.136+0,01.135=14,95\left(g\right)\)
a, Ta có: 24nMg + 56nFe = 9,2 (g) (1)
\(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
BT e, có: 2nMg + 2nFe = 2nH2 = 0,5 (2)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Mg}=0,15\left(mol\right)\\n_{Fe}=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,15.24}{9,2}.100\%\approx39,13\%\\\%m_{Fe}\approx60,87\%\end{matrix}\right.\)
b, BTNT H, có: \(n_{HCl}=2n_{H_2}=0,5\left(mol\right)\Rightarrow C_{M_{HCl}}=\dfrac{0,5}{0,2}=2,5\left(M\right)\)
