Question

How many grams of beryllium are needed to produce 36.0 g of hydrogen assume an excess of waater be(s) + 2h2o(l) be(oh)2(aq) + h2(g)?

Answer 1

Answer is: 162 g of beryllium.
Chemical reaction: Be + 2H₂O → Be(OH)₂ + H₂.
m(H₂) = 36,0 g.
m(Be) = ?
n(H₂) = m(H₂) ÷ M(H₂) 
n(H₂) = 36,0 g ÷ 2 g/mol = 18 mol.
from reaction: n(Be) : n(H₂) = 1 : 1.
n(Be) = n(H₂) = 18 mol.
m(Be) = n(Be) · M(Be).
n(Be) = 18 mol · 9 g/mol = 162 g.