\(M+2HCl\rightarrow MCl_2+H_2\left(1\right)\)
\(MCO_3+2HCl\rightarrow MCl_2+H_2O+CO_2\left(2\right)\)
Các khí trong A gồm: H2,CO2
\(n_{hhA}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
\(\overline{M}_A=11,5.2=23\left(\dfrac{g}{mol}\right)\)
Ta có: \(\overline{M}=\dfrac{M_{H_2}.n_{H_2}+M_{CO_2}.\left(0,2-n_{H_2}\right)}{0,2}\)
\(23=\dfrac{2.n_{H_2}+44\left(0,2-n_{H_2}\right)}{0,2}\)
\(\Rightarrow n_{H_2}=0,1\left(mol\right)\)
\(\Rightarrow n_{CO_2}=0,2-0,1=0,1\left(mol\right)\)
Theo (1), ta có: \(n_M=n_{H_2}=0,1\left(mol\right)\)
\(M_M=\dfrac{m_M}{n_M}=\dfrac{10,8}{0,1}=108\left(\dfrac{g}{mol}\right)\)
\(\Rightarrow M\) là Bạc (Ag)
b) Ta có: Tỉ lệ về thể tích cũng chính là tỉ lệ về số mol
\(\Rightarrow\%V_{H_2}=\dfrac{0,1}{0,2}.100\left(\%\right)=50\left(\%\right)\)
\(\Rightarrow\%V_{CO_2}=100\%-50\%=50\%\)
