Bài 1:
Gọi \(\left\{{}\begin{matrix}n_{Al}=a\left(mol\right)\\n_{Fe}=b\left(mol\right)\end{matrix}\right.\left(a,b>0\right)\)
\(n_S=\dfrac{1,28}{32}=0,04\left(mol\right)\)
PTHH:
2Al + 3S --to--> Al2S3
a---->1,5a
Fe + S --to--> FeS
b---->b
=> hệ pt \(\left\{{}\begin{matrix}27a+56b=1,1\\1,5a+b=0,04\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,02\left(mol\right)\\b=0,01\left(mol\right)\end{matrix}\right.\left(TM\right)\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,-2.27=0,54\left(g\right)\\m_{Fe}=0,01.56=0,56\left(g\right)\end{matrix}\right.\)
Bài 2:
\(n_{SO_2}=\dfrac{12,8}{64}=0,2\left(mol\right)\\ n_{KOH}=0,25.1=0,25\left(mol\right)\)
\(T=\dfrac{0,25}{0,2}=1,25\)=> Tạo cả 2 muối (KHSO3 và K2SO3)
PTHH:
2KOH + SO2 ---> K2SO3 + H2O
0,25---->0,125---->0,125
K2SO3 + SO2 + H2O ---> 2KHSO3
0,075<---0,075------------->0,15
\(\Rightarrow\left\{{}\begin{matrix}C_{M\left(K_2SO_3\right)}=\dfrac{0,125-0,075}{0,25}=0,2M\\C_{M\left(KHSO_3\right)}=\dfrac{0,15}{0,25}=0,6M\end{matrix}\right.\)
