Final answer:
In the given chemical reaction, 2Na + F2 → 2NaF, both provided scenarios have an equal amount of NaF produced from each reactant. Therefore, when both reactants are completely consumed, 3.0 moles of NaF are produced.
Step-by-step explanation:
The stoichiometric equation given is 2Na + F2 → 2NaF. According to this equation, 2 moles of sodium (Na) react with 1 mole of fluorine (F2) to produce 2 moles of sodium fluoride (NaF). The student provides two scenarios. In the first, 3.0 moles of Na can form 3.0 moles of NaF. The second scenario states that 1.4 moles of F2 can form 2.8 moles of NaF. Since the reaction is limited by the reactant that will run out first, we need to identify the limiting reagent.
From the first scenario, since 2 moles of Na produce 2 moles of NaF, then 3.0 moles of Na would indeed produce 3.0 moles of NaF. From the second scenario, it means that 1 mole of F2 produces 2 moles of NaF, hence 1.4 moles of F2 would produce 2.8 moles of NaF.
In both cases, neither of the reactants are limiting the production because they can both produce the stated amount of product. Therefore, if both reactants were present in the stated amounts, they both could produce 3.0 moles of NaF, and thus the total moles of NaF produced during the reaction would be 3.0 moles, given that all of the Na and F2 were completely consumed in the reaction.