The reaction can be written as:-\\\\(2Al_{(s)}+2NaOH_{(aq)}+2H_2O_{(l)}\\longrightarrow {2NaAlO_2}_{(aq)}+{3H_2}_{(g)}\\\\)Here, \\\\(2\\\\) moles of \\\\(Al\\\\) gives \\\\(3\\\\) moles of \\\\(H_2\\\\) gas.Moles of aluminium= \\\\(\\cfrac {0.15g}{27}=5.56\ imes 10^{-3}\\\\) moles\\\\(\ herefore\\\\) Moles of \\\\(H_2\\\\) produced= \\\\(\\cfrac {3}{2}\ imes 5.56\ imes 10^{-3}\\\\)=\\\\(8.33\ imes 10^{-3}\\\\) moles\\\\(\ herefore\\\\) Volume of \\\\(H_2\\\\)=\\\\(\\cfrac {nRT}{P}\\\\)=\\\\(\\cfrac {8.33\ imes 10^{-3}\ imes 0.0831\ imes 293}{1}\\\\)=\\\\(0.203\\\\) litres\\\\(\ herefore\\\\) Volume of dihydrogen released is \\\\(0.203\\\\) litres