Aluminium reacts with caustic soda (NaOH) in accordance with the reaction.
2Al (s) + H2O + 2NaOH → 2NaAlO2 + 2H2 (g)
2 x 27 g 3 mol
= 54 g 3 x 22.4 L
Volume of hydrogen (at STP) released when 0.15 g of Al reacts
= {0.15 g}/{54 g} x 3 x 22.4 L
= 0.187 L = 187 ml
p1 = 1 bar p2 = 1 bar
T1 = 273 K T2 = 20°C = (20 + 273)K
= 293 K
V1 = 187 mL V2 = ?
Pressure is held constant, therefore, using the relationship