Taking into account the reaction stoichiometry, 2.8 moles of NaF are formed when 3.0 mol Na reacts with 1.4 mol F₂.
Reaction stoichiometry
In first place, the balanced reaction is:
2 Na + F₂ → 2 NaF
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- Na: 2 moles
- F₂: 1 mole
- 2 NaF: 2 moles
Limiting reagent
The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.
Limiting reagent in this case
To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 2 moles of Na reacts with 1 mole of F₂, 3 moles of Na reacts with how many moles of F₂?
[tex]amount of moles of F_{2} =\frac{3 moles of Nax 1 moles of F_{2}}{2 moles of Na}[/tex]
amount of moles of F₂= 1.5 moles
But 1.4 moles of F₂ are not available, 1.4 moles are available. Since you have less moles than you need to react with 3 moles of Na, F₂ will be the limiting reagent.
Mass of NaF formed
The following rule of three can be applied: if by reaction stoichiometry 1 mole of F₂ form 2 moles of NaF, 1.4 moles of F₂ form how much mass of NaF?
[tex]moles of NaF=\frac{1.4 moles of F_{2}x2 moles of NaF }{1 mole of F_{2} }[/tex]
moles of NaF= 2.8 moles
Then, 2.8 moles of NaF are formed when 3.0 mol Na reacts with 1.4 mol F₂.
Learn more about the reaction stoichiometry:
brainly.com/question/24741074
brainly.com/question/24653699
#SPJ1
