If 23.17 g of beryllium (Be) reacts with water at standard temperature and pressure, how many liters of hydrogen (H2) are produced?

Volume of H2 produced = 57.6576 L

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

43.452

step-by-step explanation:

284 times .153 = 43.452

they are all solids.