Answer:
Step-by-step explanation:
To calculate the grams and moles of sodium oxide formed when 18.4 g of sodium reacts with oxygen, we need to determine the balanced chemical equation for the reaction between sodium and oxygen.
The balanced chemical equation for the reaction is:
4Na + O2 → 2Na2O
From the equation, we can see that 4 moles of sodium (4Na) react with 1 mole of oxygen (O2) to produce 2 moles of sodium oxide (2Na2O).
To calculate the moles of sodium oxide formed, we first need to convert the given mass of sodium to moles using its molar mass. The molar mass of sodium (Na) is approximately 22.99 g/mol.
Moles of sodium = Mass of sodium / Molar mass of sodium
Moles of sodium = 18.4 g / 22.99 g/mol ≈ 0.8 mol
Since the stoichiometry of the reaction tells us that 4 moles of sodium react to form 2 moles of sodium oxide, we can use a mole ratio to determine the moles of sodium oxide formed.
Moles of sodium oxide = (Moles of sodium / 4) * 2
Moles of sodium oxide = (0.8 mol / 4) * 2 = 0.4 mol
Finally, to calculate the mass of sodium oxide formed, we multiply the moles of sodium oxide by its molar mass. The molar mass of sodium oxide (Na2O) is approximately 61.98 g/mol.
Mass of sodium oxide = Moles of sodium oxide * Molar mass of sodium oxide
Mass of sodium oxide = 0.4 mol * 61.98 g/mol ≈ 24.79 g
Therefore, approximately 24.79 grams and 0.4 moles of sodium oxide were formed when 18.4 grams of sodium reacted with oxygen.