Answer to Question #73108 in General Chemistry for Kitty

Question

A reaction of Sodium Methoxide (NaOCH3) and Sulfuric Acid (H2SO4) was carried out resulting in methanol, which has a density of 0.792 g/mL, and Sodium Sulfate (Na2SO4) as the products. The experimenter discovered that his product was hydrated but was unsure of how many waters of hydration it had. If the experimented ended up with 0.51 mL of methanol and 4.08g of hydrated sodium sulfate, how many waters of hydration did the sodium sulfate salt have?

Answer 1

The reaction proceeds as follows:
2 NaOCH3 + H2SO4 = Na2SO4 + 2 CH3OH.
The mass of methanol is
0.51 mL (0.792 g/mL) = 0.404 g.
Converting the mass of methanol to the theoretical mass of sodium sulfate:
0.404 g CH3OH × (1 mol CH3OH / 32.04 g CH3OH) × (1 mol Na2SO4/ 2 mol CH3OH) ×
× (142.04 g Na2SO4 / 1 mol Na2SO4) = 0.896 g Na2SO4 .
The mass of water in hydrated sodium sulfate is 4.08 – 0.896 = 3.184 g, calculating the mole ratio:
moles of Na2SO4 : 0.896 g Na2SO4 × (1 mol Na2SO4/142.04 g Na2SO4 ) = 6.3 · 10-3 mol Na2SO4,
moles of H2O : 3.184 g H2O × (1 mol H2O /18.02 g H2O ) = 0.177 mol H2O.
mole ratio H2O / Na2SO4 = 0.177 / 6.3 · 10-3 = 28 moles H2O per 1 mole Na2SO4 .
So, the sodium sulfate salt have 28 waters of hydration.