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Chemistry
H2 evolved at STP on complete reaction of 27 g of aluminium with excess of aqueous NaOH would be
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Q. $H_2$ evolved at $STP$ on complete reaction of $27\, g$ of aluminium with excess of aqueous $NaOH$ would be
Some Basic Concepts of Chemistry
A
$22.4$ litres
0%
B
$44.8$ litres
100%
C
$67.2$ litres
0%
D
$33.6$ litres
0%
Solution:
$H_2O + \underset{27\,g}{Al} + NaOH \to NaAlO_2 + \underset{\frac{3}{2} \times 22.4 = 33.6\,L}{ \frac{3}{2} H_2}$