Question

12. How many grams of beryllium are needed to produce 36.0 g of hydrogen? (Assume an excess of water.)

Be(s) + 2H, O() → Be(OH)2 (aq) +H,(g)
a
4.00 g
b. 36.0 g
c. 162 g
d 324 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.

Answer 2

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.