Final answer:
The mass of sodium fluoride formed from the complete reaction of 11.5 g of fluorine with enough sodium to completely react with it is 50.82 grams.
Step-by-step explanation:
To find the mass of sodium fluoride formed from the complete reaction of 11.5 g of fluorine with enough sodium to completely react with it, we need to use stoichiometry. The balanced chemical equation for the reaction is:
2 Na + F₂ → 2 NaF
From the balanced equation, we can see that 2 moles of sodium react with 1 mole of fluorine to form 2 moles of sodium fluoride. Therefore, we need to convert the mass of fluorine to moles, and then use the stoichiometric ratio to find the mass of sodium fluoride.
First, calculate the moles of fluorine:
Moles of F = Mass of F / Molar mass of F (19 g/mol)
= 11.5 g / 19 g/mol
= 0.605 moles of F
Since the stoichiometric ratio is 2 moles of NaF to 1 mole of F, we can calculate the moles of NaF formed:
Moles of NaF = Moles of F / 1 * 2
= 0.605 moles / 1 * 2
= 1.21 moles of NaF
Finally, we can calculate the mass of NaF:
Mass of NaF = Moles of NaF * Molar mass of NaF (42 g/mol)
= 1.21 moles * 42 g/mol
= 50.82 grams