số mol H2 là:\(n=\frac{613.44}{22.4}\approx27.39\left(mol\right)\)
PTHH\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ m_{Al}=\left(27.39\cdot\frac{2}{3}\right)\cdot27=493.02\left(g\right)\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
18,25______________________27,39
\(n_{H2}=27,39\left(mol\right)\)
\(\Rightarrow m=492,75\left(g\right)\)