\(PTHH:4Al+3O_2\underrightarrow{^{to}}2Al_2O_3\left(1\right)\)
_________a_______3a/4_____0,5a
\(3Fe+2O_2\underrightarrow{^{to}}Fe_3O_4\left(2\right)\)
b_______2b/3___b/3
Gọi số nAl là a,nFe là b(a, b>0)
-Theo bài ra ta có\(27a+56b=11,1\left(3\right)\)
Theo PTHH(1) và (2) ta có
\(M_{hh}=51a+\frac{232b}{3}=16,7\left(4\right)\)
Giải hệ pt(3) và (4) ta có
\(\Rightarrow\left\{{}\begin{matrix}a=0,1\\b=0,15\end{matrix}\right.\)
Theo PTHH (1) và (2) ta có
\(\Sigma n_{O\left(pư\right)}=0,075+0,1=0,175\left(mol\right)\)
\(\Rightarrow V_{O2\left(pư\right)}=0,175.22,4=3,92\left(l\right)\)
\(\Rightarrow V_{O2\left(spu\right)}=3,92.\frac{110}{100}=4,312\left(l\right)\)
b)\(m_{Al}=0,1.27=2,7\left(g\right)\)
\(\Rightarrow\%m_{Al}=\frac{2,7.100}{11,1}=24,32\%\)
\(\Rightarrow\%m_{Fe}=100\%-24,32\%=75,68\%\)
a)\(4Al+3O2-->2Al2O3\)
x------------0,75x------------------0,5x(mol)
\(3Fe+2O2-->Fe3O4\)
y-----------2/3y---------------1/3y(mol)
Theo bài ta có hpt
\(\left\{{}\begin{matrix}27x+56y=11,1\\51x+\frac{232}{3}y=16,7\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,15\end{matrix}\right.\)
\(n_{O2\left(1\right)}=\frac{3}{4}n_{Al}=0,075\left(mol\right)\)
\(n_{O2\left(2\right)}=\frac{2}{3}n_{Fe}=0,1\left(mol\right)\)
\(\sum n_{O2}=0,175\left(mol\right)\)
\(V_{O2}=0,175.22,4=3,92\left(l\right)\)
b)\(\%m_{Al}=\frac{0,1.27}{11,1}.100\%=24,32\%\)
\(\%m_{Fe}=100-24,32=75,68\%\)
