yield = 56.95 %
Explanation:
We have the following chemical reaction:
4 Na (s) + O₂ (g) → 2 Na₂O (s)
number of moles = mass / molecular weight
number of moles of Na = 1.72 × 10⁻³ / 23 = 0.0748 × 10⁻³ moles
number of moles of Na₂O = 1.32 × 10⁻³ / 62 = 0.0213 × 10⁻³ moles (practical)
Taking in account the stoichiometry of the chemical reaction, we devise the following reasoning:
if 4 moles of Na produces 2 moles of Na₂O
then 0.0748 × 10⁻³ moles of Na produces X moles of Na₂O
X = (0.0748 × 10⁻³ × 2) / 4 = 0.0374 × 10⁻³ moles of Na₂O (theoretical)
yield = (practical mass / theoretical mass) × 100
yield = (0.0213 × 10⁻³ / 0.0374 × 10⁻³) × 100
yield = 56.95 %
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