Solution
Correct option is A)
2Al+3H
2
SO
4
→Al
2
(SO
4
)
3
+3H
2
27
5.7
1000
50×5
=0.2mole =0.25 mole (Limiting reagent)
∴0.25 mole H
2
is formed
PV=nRT
V=
P
nRT
=
1atm
0.25×0.082×300
V=6.15 litre