Question

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

Answer 1

Volume of Sodium metal : 20.454 ml

Further explanation

Reaction(balanced) :

2Na + 2H₂O → 2NaOH + H₂

mass NaOH = 34.5 g

mol NaOH(MW=40 g/mol) :

`\tt mol=(mass)/(MW)\\\\mol=(34.5)/(40)\\\\mol=0.8625`

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

Mass Na (Ar=23 g/mol):

`\tt mass=mol* MW\\\\mass=0.8625* 23\\\\mass=19.84~g`

Volume Na :

`\tt V=(m)/(\rho)\\\\V=(19.84~g)/(0.97~g/ml)\\\\V=20.454~ml`