\(n_{H_2}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\) ( p.ứ thế )
0,3 0,45 0,15 0,45 ( mol )
\(m_{Al}=0,3.27=8,1\left(g\right)\)
\(m_{H_2SO_4}=0,45.98=44,1\left(g\right)\)
\(C1.m_{Al_2\left(SO_4\right)_3}=0,15.342=51,3\left(g\right)\)
\(C2.BTKL:m_{Al}+m_{H_2SO_4}=m_{Al_2\left(SO_4\right)_3}+m_{H_2}\)
\(\Leftrightarrow8,1+44,1=m_{Al_2\left(SO_4\right)_3}+0,45.2\)
\(\Leftrightarrow m_{Al_2\left(SO_4\right)_3}=51,3\left(g\right)\)