a.; b. Theo đề, ta có: \(n_{H_2S}=\frac{2,24}{22,4}=0,1\left(mol\right);n_{KOH}=0,2.1=0,2\left(mol\right)\)
Ta có tỉ lệ: \(T=\frac{n_{KOH}}{n_{H_2S}}=\frac{0,2}{0,1}=2\)
\(\Rightarrow\) Sau phản ứng tạo muối \(K_2S\).
PTHH: \(H_2S+2KOH\rightarrow K_2S+2H_2O\)
Mol: \(0,1------>0,1\)
Theo phương trình: \(n_{K_2S}=n_{H_2S}=0,1\left(mol\right)\Rightarrow m_{K_2S}=0,1.110=11\left(g\right)\)
a)
\(\left\{{}\begin{matrix}nKOH=1,5\times0.1=0.15\left(mol\right)\\nH_2S=\frac{2,24}{22,4}=0,1\left(mol\right)\\\end{matrix}\right.\)
T=\(\frac{nKOH}{nH_2S}=\frac{0,15}{0,1}=1,5\)
⇒Sp:\(\left\{{}\begin{matrix}KHS\\K_2S\end{matrix}\right.\)
b)
Ptpư:
\(H_{2_{_x}}S+KO_{_{_x}}H\rightarrow KH_{_{_x}}S+H_{2_{_x}}O\)
\(H_{2_{_y}}S+2KO_{_{_{2y}}}H\rightarrow K_{2_{_y}}S+H_{2_{_y}}O\)
ta có hệ phương trình:
\(\left\{{}\begin{matrix}x+y=0,1\\x+2y=0,15\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,05\\y=0,05\end{matrix}\right.\)
\(\Rightarrow mKHS=0,05\times\left(39+1+32\right)=3,6\)
\(\Rightarrow K_2S=0,05\times\left(39\times2+32\right)=5,5\)
\(m\)Muối=3,6+5,5=9,1
