Correct Answer - C
The reaction between aluminium and caustic soda is
`underset(2xx27=54g)(2Al)+2NaOH+2H_(2)O rarr 2NaAlO_(2)+underset(3xx22.4" L at STP")(3H_(2))`
`therefore 54 g` of Al produces `H_(2)` at `STP = 3xx22.4 L`
0.15 g of Al will produce `H_(2)` at STP
`=(3xx22.4)/(54)xx0.15=0.186 L`
`{:("At STP","Given conditions"),(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):}`
Applying ideal gas rquation, `(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))`
`(1xx0.1867)/(273)=(0.987xx V_(2))/(293)`
`V_(2)=(293)/(0.987)xx(1xx0.1867)/(273)=0.2030 L = 203 mL` .