Notice how the K and Ag are both being swapped around.
Single Replacement:
A+BX → B+AX
Double Replacement:
AX+BY → BX + AY
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.
Answer:
The empirical formula of the compound C₃H₆N₂ is C₃H₆N₂
Explanation:
The empirical formula of a compound is the formula of the compound given in the (smallest) whole number ratio of the elements of the compound
The empirical formula of S₂O₂ is SO
The empirical formula of C₃H₆O₃ is CH₂O
The given compound's molecular formula is C₃H₆N₂
The smallest whole number ratio of of the elements of the compound is 3:6:2, therefore, the empirical formula of the compound C₃H₆N₂ is C₃H₆N₂.