The balanced chemical equation for the reaction of sodium with oxygen is:
4Na + O2 -> 2Na2O
From the balanced equation, we can see that 4 moles of sodium react with 1 mole of O2 to produce 2 moles of Na2O.
Now, let's calculate the molar mass of Na2O:
Molar mass of Na = 22.99 g/mol (sodium)
Molar mass of O = 16.00 g/mol (oxygen)
Molar mass of Na2O = 2(Molar mass of Na) + Molar mass of O
= 2(22.99 g/mol) + 16.00 g/mol
= 45.98 g/mol + 16.00 g/mol
≈ 61.98 g/mol
Now, let's calculate the moles of Na2O produced from 184 g of sodium:
Moles of Na2O = Mass of sodium / Molar mass of Na2O
= 184 g / 61.98 g/mol
≈ 2.970 moles of Na2O
Since the mole ratio of Na2O to O2 is 2:1 in the balanced equation, we need half as many moles of O2:
Moles of O2 needed = 0.5 * Moles of Na2O
≈ 0.5 * 2.970 moles
≈ 1.485 moles of O2
Now, let's calculate the mass of oxygen needed:
Mass of O2 needed = Moles of O2 needed * Molar mass of O2
≈ 1.485 moles * 32.00 g/mol (molar mass of O2)
≈ 47.52 g of O2 (approximately)
So, approximately 47.52 grams of oxygen is needed to react with 184 grams of sodium.