Respuesta :
Answer:
25%
Explanation:
In this case, let's write the reaction again:
Xe(g) + 2F₂(g) -------> XeF₄(g)
Now, we know that we got 0.25 mol of the product. We want to know the percent yield. To get this value, we need to calculate the theorical yield.
We have 2.2 mol of Xe and 2 mol of F₂, we need to know the limiting reactant.
According to the reaction, the mole ratio:
1 moles Xe --------> 2 moles F₂
According to this, and the given data, we have 2.2 moles of Xe, so:
1 moles Xe --------> 2 moles F₂
2.2 moles Xe ----------> X
X = 2.2 * 2 / 1 = 4.4 moles of F₂
However we only have 2 moles of F₂, therefore, the limiting reactant is F₂ while the excess is the Xe.
So the actual moles reacting here would be 2 moles of F₂, and this will produce X moles of XeF₄:
2 mole of F₂ --------> 1 mole XeF₄
The moles of XeF₄ would be 1 mole theorical.
If we got 0.25 then the percent yield would have to be:
% = 0.25 / 1 * 100
% = 25%
The percentage yield of XeF from the concentration of the given reactants will be 25%.
The reaction states that 1 mole of Xe will give 1 mole of XeF, and 2 moles of fluorine, will gives 1 mole of XeF.
The limiting reactant can be calculated:
1 mole Xe = 2 moles of [tex]\rm F_2[/tex]
2.2 moles of Xe = 2.2 [tex]\times[/tex] 2 moles of [tex]\rm F_2[/tex]
2.2 moles of Xe = 4.4 moles of [tex]\rm F_2[/tex]
Since the amount of available [tex]\rm F_2[/tex] has been in the limiting, thus [tex]\rm F_2[/tex] has been the limiting reactant.
So, the yield of XeF in terms of [tex]\rm F_2[/tex] will be:
2 moles [tex]\rm F_2[/tex] = 1 mole XeF
Thus the theoretical yield of XeF is 1 mole.
The yield of XeF we get = 0.25 moles.
Thus the percentage yield = [tex]\rm \dfrac{Obtained}{Theoretical\;yield}\;\times\;100[/tex]
Percentage yield = [tex]\rm \dfrac{0.25}{1}\;\times\;100[/tex]
Percentage yield = 25%
Thus the percentage yield of XeF from the concentration of the given reactants will be 25%.
For more information about the percent yield, refer to the link:
https://brainly.com/question/12809634
