Ta có:
\(n_{hh}=\frac{2,24}{22,4}=0,1\left(mol\right)\)
Gọi \(\left\{{}\begin{matrix}n_{SO2}:x\left(mol\right)\\n_{H2S}:y\left(mol\right)\end{matrix}\right.\)
\(M_{hh}=24,5.2=49\left(\frac{g}{mol}\right)\)
\(\Rightarrow m_{hh}=0,1.49=4,9\left(g\right)\)
Giải hệ PT:
\(\left\{{}\begin{matrix}64x+34y=4,9\\x+y=0,1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,05\\y=0,05\end{matrix}\right.\)
Bảo toàn e:
\(n_R=\frac{0,5}{n}\left(mol\right)\Rightarrow M_R=\frac{4,5}{\frac{0,5}{n}}=9n\left(\frac{g}{mol}\right)\)
Biện luận :
\(n=3\Rightarrow M_R=27\left(\frac{g}{mol}\right)\)
Vậy R là nhôm ( Al )
\(\Rightarrow m=m_{Al}+96.\left(n_{SO2}+4n_{H2S}\right)=28,5\left(g\right)\)