Question

If you have 12.5g of fluoride and 16.2g of sodium, which is the limiting reactant and how sodium fluoride in grams is your theoretical yield?
Na + F2 ------> NaF

Answer 1

Answer:

F2 is the limiting reactant

27.6 grams of NaF is produced.

Explanation:

Balance the equation first.

2Na+ F2 ---> 2NaF

To find the limiting reactant, solve for how much NaF can be produced with Na and F2

12.5g F2 x (1 mole F2/ 38.00 grams F2)x (2 mole NaF/ 1 mole F2)

=0.658 moles NaF

16.2g Na x (1 mole Na/ 22.99 grams Na)x (2 mole NaF/ 2 mole Na)

=0.705 moles NaF

Since F2 produced the least NaF, F2 is the limiting reactant.

Now, to find how much NaF there is, use the moles solved above with F2 as the limiting reactant.

0.658 moles NaF x (41.99 grams NaF/ 1 mole NaF)= 27.6 moles NaF

27.6 moles of NaF would be theoretically produced.