\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)
\(2Al\left(OH\right)_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+6H_2O\)
\(m_{Al_2\left(SO_4\right)_3}=\dfrac{273,75.21,863}{100}=59,85\left(g\right)\)
\(\Rightarrow n_{Al_2\left(SO_4\right)_3}=0,175\left(mol\right)\)
\(\Rightarrow n_{Al}\)(trong hỗn hợp ban đầu) \(=0175.2=0,35\left(mol\right)\)
Theo PTHH , ta có: \(n_{H_2SO_4}=\dfrac{3}{2}.n_{Al}\left(trong.hh.bđ\right)=\dfrac{3}{2}.0,35=0,525\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=51,45\left(g\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{51,45.100}{20}=257,25\left(g\right)\)
\(n_{H_2}=\dfrac{5,04}{22,4}=0,225\left(mol\right)\Rightarrow m_{H_2}=0,45\left(g\right)\)
Ta có: \(m+m_{ddH_2SO_4}=m_{ddAl_2\left(SO_4\right)_3}+m_{H_2}\)
\(\Leftrightarrow m+257,25=273,75+0,45\)
\(\Rightarrow m=16,95\left(g\right)\)
