Final answer:
When 41.0 g of Na reacts, it would theoretically produce approximately 55.3 g of Na₂O, given a 4:1 reaction stoichiometry with oxygen. None of the answer choices provided (76.0 g, 86.0 g, 94.0 g, 110.0 g) are correct.
Step-by-step explanation:
To determine the mass of Na₂O produced when 41.0 g of Na reacts, we need to follow the stoichiometry of the reaction between sodium (Na) and oxygen to form sodium oxide (Na₂O). Unfortunately, the chemical equation for this reaction is not provided. However, typically sodium reacts with oxygen in a 4:1 ratio to form Na₂O:
4Na + O₂ → 2Na₂O.
Using this equation, we calculate the moles of Na: 41.0 g Na * (1 mol Na/22.99 g Na) = 1.7839 mol Na.
According to the reaction stoichiometry, 4 moles of Na produce 2 moles of Na₂O. Therefore, the moles of Na₂
produced are (1.7839 mol Na) * (2 mol Na₂O / 4 mol Na) = 0.89195 mol Na₂O.
Now, converting moles of Na₂O to grams: 0.89195 mol Na₂O * (61.98 g Na₂O/mol) = 55.30 g Na₂O.
All the answer choices are incorrect based on the provided molar mass and typical reaction stoichiometry. The correct mass of Na₂O produced from 41.0 g of Na should be approximately 55.3 g, not one of the provided options.