Question

If 23.17 g of beryllium (Be) reacts with water at standard temperature and pressure,

Answer 1

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