Ta có:
\(n_{Al}=\frac{2,7}{27}=0,1\left(mol\right)\)
Phản ứng xảy ra:
\(2Al+6HX\rightarrow2AlX_3+3H_2\)
\(n_{H2}=\frac{3}{2}n_{Al}=0,15\left(mol\right)\)
BTKL: \(m_{dd.A}=m_{Al}+m_X=-m_{H2}=2,7+50-0,15.2=52,4\left(g\right)\)
\(\Rightarrow m_{AlX3}=50,95\%.52,4=26,7\left(g\right)\)
Ta có:
\(n_{AlX3}=n_{Al}=0,1\left(mol\right)\)
\(\Rightarrow m_{AlX3}=27+3X=\frac{26,7}{1}=267\)
\(\Rightarrow X=80\left(Br\right)\)
\(\Rightarrow n_{HBr}=3n_{Al}=0,3\left(mol\right)\)
\(AgNO_3+HBr\rightarrow AgBr+HNO_3\)
\(\Rightarrow n_{AgBr}=n_{HBr}=0,3\left(mol\right)\)
\(\Rightarrow m_{AgBr}=0,3.\left(108+80\right)=56,4\left(g\right)\)