Volume of H2 produced = 57.6576 L
Further explanation
Given
23.17 g Be
Required
Volume of H2
Solution
Reaction
Be(s)+H2O(g)→BeO(s)+H2(g)
mol Be :
= 23.17 g : 9 g/mol
= 2.574
From the equation, mol H2 : mol Be = 1 : 1, so mol H2 = 2.574
Volume H2(assumed at STP, 1 mol=22.4 L) :
= 2.574 x 22.4 L
= 57.6576 L