`underset(2xx27=54)(2Al)+2NaOH+2H_(2)O rarr 2NaAlO_(2)+underset(3xx22.4 L)(3H_(2))`
`54 g Al` gives `3xx22.4 L H_(2)`
`0.15 g Al` gives `=(0.15xx3xx22.4)/(54)=0.186 L`
`(P_(1)V_(1))/T_(1)=(P_(2)V_(2))/T_(2)`
`P_(1)=1 atm," " P_(2)=1 "bar" =0.987 atm`
`V_(1)=0.186 L," " V_(2)=?`
`T_(1)=273 K," " T_(2)=273+20=293 K`
`(1xx0.186)/273=(0.987xxV_(2))/293`
or `V_(2)=(1xx0.186xx293)/(0.987xx237)=0.202 L=202 mL`