Taking into account definition of percent yield, the percent yield for the reaction is 100%.
Reaction stoichiometry
In first place, the balanced reaction is:
2 FeBr₃ + 3 Na₂S → Fе₂S₃ + 6 NaBr
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:
- FeBr₃: 2 moles
- Na₂S; 3 moles
- Fе₂S₃: 1 mole
- NaBr: 6 moles
Moles of NaBr formed
The following rule of three can be applied: if by reaction stoichiometry 2 moles of FeBr₃ form 6 moles of NaBr, 2.36 moles of FeBr₃ form how many moles of NaBr?
[tex]moles of NaBr=\frac{2.36 moles of FeBr_{3}x6 moles of NaBr }{2 moles of FeBr_{3}}[/tex]
moles of NaBr= 7.08 moles
Then, 7.08 moles of NaBr can be produced from 2.36 moles of FeBr₃.
Percent yield
The percent yield is the ratio of the actual return to the theoretical return expressed as a percentage.
The percent yield is calculated as the experimental yield divided by the theoretical yield multiplied by 100%:
[tex]percent yield=\frac{actual yield}{theorical yield}x100[/tex]
where the theoretical yield is the amount of product acquired through the complete conversion of all reagents in the final product, that is, it is the maximum amount of product that could be formed from the given amounts of reagents.
Percent yield for the reaction in this case
In this case, being the molar mass of NaBr 102.9 g/mole, you know:
- actual yield= 7.08 moles× 102.9 g/mole= 728.532 grams
- theorical yield= 7.08 moles× 102.9 g/mole= 728.532 grams
Replacing in the definition of percent yields:
[tex]percent yield=\frac{728.532 grams}{728.532 grams}x100[/tex]
Solving:
percent yield= 100%
Finally, the percent yield for the reaction is 100%.
Learn more about
the reaction stoichiometry:
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percent yield:
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