What is the theoretical yield of beryllium needed to produce 36.0 g of hydrogen? (Assume an excess of water.)
Be(s) + 2H2O(l) Be(OH)2 (aq) + H2(g)

a. 4.00 g
b. 36.0 g
c. 162 g
d. 324 g

Question

What is the theoretical yield of beryllium needed to produce 36.0 g of hydrogen? (Assume an excess of water.)
Be(s) + 2H2O(l)  Be(OH)2 (aq) + H2(g)

a. 4.00 g
b. 36.0 g
c. 162 g
d. 324 g

anonymous · Answer

Moles H2 = 36.0 g / 2.016 g/mol= 17.9
the ratio between Be and H2 is 1 : 1
moles Be needed = 17.9
Mass Be = 17.9 mol x 9.01218 g/mol=161.3 g

anonymous · Answer

how about this one: 
TRUE OR FALSE: Mole ratios in stoichiometry are used to convert one substance to find a quantity of another substance using an unbalanced chemical equation.