a)\(PTHH:2Zn+O_2\rightarrow2ZnO\)
__________a____0,5a__________
\(4Al+3O_2\rightarrow2Al_2O_3\)
b___ 0,75b________
Gọi nZn là a,nAl là b(a,b>0)
Theo bài ra ta có
\(65a+27b=18,4\left(1\right)\)
\(n_{O2}=0,5a+0,76b\left(mol\right)\)
\(n_{O2}=\frac{5,6}{22,4}=0,25\left(mol\right)\)
\(\Rightarrow0,5a+0,75b=0,25\left(2\right)\)
Giải hệ PT (1) và (2)
\(\Rightarrow\left\{{}\begin{matrix}a=0,2\left(mol\right)\\b=0,2\left(mol\right)\end{matrix}\right.\)
\(m_{Zn}=0,2.65=13\left(g\right)\)
\(m_{Al}=0,2.27=5,4\left(g\right)\)
\(\Rightarrow\%m_{Zn}=\frac{13}{18,4}.100\%=70,65\%\)
\(\Rightarrow\%m_{Al}=100\%-70,65\%=29,35\%\)
b)\(PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\)
_________0,1___________0,1__________
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,1_____________________0,15
Ta có: \(\left\{{}\begin{matrix}n_{Zn\left(b\right)}=0,1\left(mol\right)\\n_{Al\left(b\right)}=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\Sigma n_{H2}=0,25\left(mol\right)\)
\(\Rightarrow V_{H2}=0,25.22,4=5,6\left(l\right)\)
