TH1:
\(2R+2nH_2SO_4\rightarrow R_2\left(SO_4\right)_n+nSO_2+2nH_2O\)
\(\frac{n_R}{n_{H2SO4}}=\frac{1}{1,25}\Rightarrow\frac{1}{n}=\frac{1}{1,25}\)
\(\Rightarrow n=1,25\)
\(\Rightarrow2R+2,5H_2SO_4\rightarrow R_2\left(SO_4\right)_{1,25}+1,25SO_2+2,5H_2O\)
\(n_{Br2}=0,1\left(mol\right)\)
\(SO_2+Br_2+2H_2O\rightarrow2HBr+H_2SO_4\)
0,1____0,1______________________________
\(n_{R2\left(SO4\right)1,5}=0,08\left(mol\right)\)
\(\Rightarrow M_{R2\left(SO4\right)1,5}=\frac{12}{0,08}=150=2R+1,5.96\)
\(\Rightarrow R=3\) (loại)
TH2 : \(8R+5nH_2SO_4\rightarrow4R_2\left(SO_4\right)_n+nH_2S+4nH_2O\)
\(\frac{n_2}{n_{H2SO4}}=\frac{1}{1,25}=\frac{4}{5}\Rightarrow\frac{8}{5n}=\frac{4}{5}\)
\(\Rightarrow n=2\)
\(4R+5H_2SO_4\rightarrow4RSO_4+H_2S+4H_2O\)
\(\Rightarrow n_{Br2}=0,1\left(mol\right)\)
\(H_2S+4Br_2+4H_2O\rightarrow8HBr+H_2SO_4\)
0,025___0,1___________________________
\(\Rightarrow n_{RSO4}=0,1\left(mol\right)\)
\(\Rightarrow M_{RSO4}=\frac{12}{0,1}=120=R+96\)
\(\Rightarrow R=24\left(Mg\right)\)
Vậy kim loại R là Magie (Mg)
