Gọi số mol Al là x; Zn là y.
\(\Rightarrow27x+65y=2,49\)
Gọi số mol Al là x; Zn là y.
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
Ta có:
\(n_{H2}=\frac{3}{2}n_{Al}+n_{Zn}=1,5x+y=\frac{1,344}{22,4}=0,0\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}x=0,02\\y=0,03\end{matrix}\right.\)
\(\Rightarrow m_{Al}=0,02.27=0,54\left(g\right)\)
\(\Rightarrow\%m_{Al}=\frac{0,54}{2,49}=21,49\%\Rightarrow\%m_{Zn}=78,31\%\)
\(n_{H2SO4}=n_{H2}=0,06\left(mol\right)\)
BTKL:
\(m_{kl}+m_{H2SO4}=m_{muoi}+m_{H2}\)
\(\Rightarrow m_{muoi}=2,49+0,06.98-0,06.2=8,25\left(g\right)\)
Cho hỗn hợp trên tác dụng với H2SO4 đặc nguội thì Al bị thụ động hóa, chỉ có Zn phản ứng.
\(Zn+H_2SO_{4\left(đac\right)}\rightarrow ZnSO_4+SO_2+2H_2O\)
\(\Rightarrow n_{SO2}=n_{Zn}=0,03\left(mol\right)\)
\(\Rightarrow V_{SO2}=0,03.22,4=0,672\left(l\right)\)
a) \(n_{H2}=\frac{1,344}{22,4}=0,06\left(mol\right)\)
\(2Al+3H2SO4-->Al2\left(SO4\right)3+3H2\)
x-----------------------------------------------1,5x(mol)
\(Zn+H2SO4-->ZnSO4+H2\)
y-------------------------------------------y(mol)
Theo bài ta có hpt
\(\left\{{}\begin{matrix}27x+65y=2,49\\1,5x+y=0,06\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,02\\y=0,03\end{matrix}\right.\)
\(\%m_{Al}=\frac{0,02.27}{2,49}.100\%=21,69\%\)
\(\%m_{Zn}=100-21,69=78,31\%\)
b) \(m_{muối}=m_{Al2\left(SO4\right)3}+m_{ZnSO4}=0,02.400+0,03.161=12,83\left(g\right)\)
c)\(2Al+6H2SO4-->Al2\left(SO4\right)3+6H2O+3SO2\)
0,02----------------------------------------------------------0,03(mol)
\(Zn+2H2SO4-->ZnSO4+2H2O+SO2\)
0,03------------------------------------------------------0,03(mol)
\(\sum n_{SO2}=0,03+0,03=0,06\left(mol\right)\)
\(V_{SO2}=0,06.22,4=1,344\left(l\right)\)