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a) Fe + 2HCl --> FeCl2 + H2
b) \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: Fe + 2HCl --> FeCl2 + H2
_____0,15<-0,3<----0,15<---0,15
\(\left\{{}\begin{matrix}\%Fe=\dfrac{0,15.56}{12}.100\%=70\%\%\\\%Cu=100\%-70\%=30\%\end{matrix}\right.\)
c) mHCl = 0,3.36,5 = 10,95 (g)
=> \(m_{dd}=\dfrac{10,95.100}{10}=109,5\left(g\right)\)
d) mdd = 12 + 109,5 - 0,15.2 = 121,2 (g)
\(C\%\left(FeCl_2\right)=\dfrac{0,15.127}{121,2}.100\%=15,718\%\)
a)
$Fe + 2HCl \to FeCl_2 + H_2$
$FeO + 2HCl \to FeCl_2 + H_2O$
b)
Theo PTHH : $n_{Fe} = n_{H_2} = \dfrac{3,36}{22,4} = 0,15(mol)$
$m_{Fe} = 0,15.56 = 8,4(gam)$
$m_{FeO} = 12 - 8,4 = 3,6(gam)$
$n_{FeO} =0,05(mol)$
Theo PTHH : $n_{HCl} = 2n_{Fe} + 2n_{FeO} = 0,4(mol)$
$V_{dd\ HCl} = \dfrac{0,4}{2} = 0,2(lít)$
c) $Fe + CuSO_4 \to FeSO_4 + Cu$
$n_{Cu} = n_{Fe} = 0,15(mol) \Rightarrow m_{chất\ rắn} = m_{FeO} + m_{Cu}$
$= 3,6 + 0,15.64 = 13,2(gam)$
a) PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
a_____________________\(\dfrac{3}{2}\)a (mol)
\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
b____________________b (mol)
Ta lập HPT: \(\left\{{}\begin{matrix}27a+56b=11\\\dfrac{3}{2}a+b=\dfrac{8,96}{22,4}=0,4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,2\cdot27}{11}\cdot100\%\approx49,09\%\\\%m_{Fe}=50,91\%\end{matrix}\right.\)
b) PTHH: \(CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\)
Ta có: \(\left\{{}\begin{matrix}n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\\n_{H_2}=\dfrac{3}{2}a+b=0,4\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) H2 còn dư, tính theo CuO
\(\Rightarrow n_{Cu}=0,2\left(mol\right)\) \(\Rightarrow m_{Cu}=0,2\cdot64=12,8\left(g\right)\)
Gọi n Al = a ( mol ) , n Fe = b ( mol )
Có: n H2 = 0,4 ( mol )
PTHH
2AL + 6HCL ===> 2ALCL3 + 3H2
a--------------------------------------a
Fe + 2HCl ====> FeCL2 + H2
b------------------------------------b
Ta có hpt:
\(\left\{{}\begin{matrix}27a+56b=11\\1,5a+b=0,4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\)
=> m AL = 5,4 ( g ) ; m Fe = 5,6 ( g )
b) Có : n CuO = 0,2 ( mol )
PTHH:
CuO + H2 ====> Cu +H2O
0,2----0,2-----------0,2
theo pthh: n Cu = 0,2 ( mol ) => m Cu = 12,8 ( g )
a, \(Fe+2HCl\rightarrow FeCl_2+H_2\)
b, \(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
Theo PT: \(n_{Fe}=n_{H_2}=0,4\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,4.56}{35}.100\%=64\%\\\%m_{Cu}=36\%\end{matrix}\right.\)
a, \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
PTHH: Zn + 2HCl → ZnCl2 + H2
Mol: 0,2 0,2
(do Cu ko tác dụng với HCl loãng)
b, \(m_{Zn}=0,2.65=13\left(g\right)\)
\(m_{Cu}=19,4-13=6,4\left(g\right)\)
a)
$Fe + 2HCl \to FeCl_2 + H_2$
b)
Theo PTHH : $n_{Fe} = n_{H_2} = \dfrac{4,48}{22,4} = 0,2(mol)$
$\Rightarrow m_{Fe} = 0,2.56 = 11,2(gam)$
$\Rightarrow m_{Cu} = 15,6 - 11,2 = 4,4(gam)$