Reaction of alcohols with alkali metals

Question

When 106 grams of a mixture of cyclopentanol and water reacts with excess Na, 44.8 L of H2is produced atSTP. Find the mole percentage of cyclopentanol in the mixture?

Answer 1

2Na(s) + 2H2O ==> 2NaOH + H2(g)

At STP 44.8 L x 1 mol/22.4 L = 2 moles H2 produced

2 mole H2 x 2 mol H2O/mol H2 = 4 mol H2O used

moles cyclopentanol present = 106 g x 1 mol/86.1 g = 1.23 moles. 

Total moles = 4  + 1.23 = 5.23 moles

mole percent cyclopentanol = 1.23 mol/5.23 mol (x100%) = 23.5%