a, \(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
PTHH: 2Al + 6HCl → 2AlCl3 + 3H2
Mol: x 1,5x
PTHH: Fe + 2HCl → FeCl2 + H2
Mol: y y
b, Ta có hpt: \(\left\{{}\begin{matrix}27x+56y=11\\1,5x+y=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)
\(\Rightarrow\%m_{Al}=\dfrac{0,2.27.100\%}{11}=49,09\%\Rightarrow\%m_{Fe}=100\%-49,09\%=50,91\%\)
c, \(n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\)
Ta có: \(\dfrac{0,2}{1}< \dfrac{0,4}{1}\) ⇒ CuO hết, H2 dư
PTHH: CuO + H2 → Cu + H2O
Mol: 0,2 0,2
\(m_{Cu}=0,2.64=12,8\left(g\right)\)
Gọi \(m_{Al}=a\left(g\right)\left(0< a< 11\right)\)
\(\rightarrow m_{Fe}=11-a\left(g\right)\)
\(\rightarrow\left\{{}\begin{matrix}n_{Al}=\dfrac{a}{27}\left(mol\right)\\n_{Fe}=\dfrac{11-a}{56}\left(mol\right)\end{matrix}\right.\)
PTHH:
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(\dfrac{a}{27}\) \(\dfrac{a}{18}\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(\dfrac{11-a}{56}\) \(\dfrac{11-a}{56}\)
\(\rightarrow pt:\dfrac{a}{18}+\dfrac{11-a}{56}=0,4\\ \Leftrightarrow m_{Al}=a=5,4\left(g\right)\left(TM\right)\)
\(\rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{5,4}{11}=49,1\%\\\%m_{Fe}=100\%-49,1\%=50,9\%\end{matrix}\right.\)
\(n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\)
PTHH: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
LTL: \(0,2< 0,4\rightarrow\) H2 dư
\(n_{Cu}=n_{CuO}=0,2\left(mol\right)\rightarrow m_{CuO}=0,2.64=12,8\left(g\right)\)
a. \(n_{H_2}=\dfrac{8.96}{22,4}=0,4\left(mol\right)\)
PTHH : 2Al + 6HCl -> 2AlCl3 + 3H2
\(\dfrac{0.8}{3}\) 0,4
Fe + 2HCl -> FeCl2 + H2
\(m_{Al}=\dfrac{0,8.27}{3}=7,2\left(g\right)\)
\(m_{Fe}=11-7,2=3,8\left(g\right)\)
b. \(\%m_{Al}=\dfrac{7.2}{11}.100=65,5\%\)
\(\%m_{Fe}=100\%-65,5\%=34,5\%\)
c. \(n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\)
PTHH : H2 + CuO ----to----> Cu + H2O
0,2 0,2
\(m_{Cu}=0,2.64=12,8\left(g\right)\)