How many milliliters of sodium metal, with a density of 0.97 g/mL, would be needed to produce 34.5 grams of sodium hydroxide in the single replacement reaction below? Unbalanced equation: Na + H2O ---> NaOH + H2

Volume of Sodium metal : 20.454 ml

Reaction(balanced) :

2Na + 2H₂O → 2NaOH + H₂

mass NaOH = 34.5 g

mol NaOH(MW=40 g/mol) :

From the equation, mol ratio of Na : NaOH = 2 : 2, so mol Na=mol NaOH=0.8625

Mass Na (Ar=23 g/mol):

Volume Na :

fe 55.85 g/mol and h2o 18.02 g/mol.