Final answer:
When 32.5 g of sodium reacts with oxygen, approximately 0.7065 moles of sodium peroxide are produced, based on the stoichiometry of the reaction which compares the mole ratio between sodium and sodium peroxide.
Step-by-step explanation:
To calculate the moles of sodium peroxide (Na2O2) produced, we first need to determine the molar mass of sodium (Na), which is approximately 23.0 g/mol. When 32.5 g of Na reacts, we can find the moles of Na that reacted by dividing the mass by its molar mass.
Number of moles of Na = 32.5 g ÷ 23.0 g/mol = 1.413 moles of Na (approximately)
The reaction of sodium with oxygen to form sodium peroxide is as follows:
4Na(s) + O2(g) → 2Na2O2(s)
According to the balanced chemical equation, 4 moles of Na react with 1 mole of O2 to produce 2 moles of Na2O2. Since we have 1.413 moles of Na, we can calculate the moles of Na2O2 produced by using the stoichiometry of the reaction:
1.413 moles Na × (1 mole Na2O2 / 2 moles Na) = 0.7065 moles Na2O2
So, 32.5 g of sodium, when reacted with oxygen, will produce approximately 0.7065 moles of sodium peroxide.