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: 19.4 g/cm3
Explanation: density is the relationship between mass over volume.
So density of gold is 15.7g/0.81cm3 = 19.4 g/cm3
Well, we need to find the ratio of Al to the other reactant.
Al:HCl = 1:3
--> this means that for every 1 Al used, you have to use 3 HCl.
6*3 = 18 moles of HCl needed to fully react with 6 moles of Al. Since 13<18, HCL is the limiting reactant.
The ratio of HCl:AlCl = 3:1
13/3 = 4.3333...
The final answer is HCl is the limiting reactant with 4.3 moles of AlCl3 able to be produced.
Hope this helps!!! :)
