How many grams of beryllium are needed to produce 11.5 g of hydrogen gas? Be(s) + 2H2O(l) → Be(OH)2(aq) + H2(g)?

Molar mass of Be = 9.012 g/mol
Molar mass of H₂ = 1.008× 2 g/mol = 2.016 g/mol

Be(s) + 2H₂O(ℓ) → Be(OH)₂(aq) + H₂(g)
Mole ratio Be : H₂ = 1 : 1

Moles of H₂ produced = (11.5 g) / (2.016 g/mol) = 5.70 mol
Moles of Be needed = 5.70 mol
Mass of Be needed = (5.70 mol) × (9.012 g/mol) = 51.4 g

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OR:
(11.5 g H₂) × (1 mol H₂ / 2.016 g H₂) × (1 mol Be / 1 mol H₂) × (9.012 g Be / 1 mol Be)
= 51.4 g Be

Be(s) + 2H2O(l) → Be(OH)2(aq) + H2(g)

moles of Be = 11.5g/atomic mass = 11.5/9.012 = 1.276

from above equation ,

1 mole of Be (s) gives---> 1 mole of H2(g)

1.276 mole of Be (s) gives---> 1.276 mole of H2(g)

Now, 1 mole of H2(g) --> 2.0158

 1.276 mole of H2(g)------>  1.276 x  2.0158 = 2.572g

  .
.   .  mass of H2(g) -->  2.572g

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