Determine the percent yield for the reaction between 46.5 g of
ZnS and 13.3 g of oxygen if 18.l4 g of ZnO is recovered along with an unknown quantity of sulfur
dioxide.
please show work thank you
Answers
The percentage yield of the reaction has been 46.74%.
Percentage yield can be given as the ratio of the actual yield to the theoretical yield.
For the theoretical yield, the balanced equation will be:
[tex]\rm 2\;ZnS\;+\;3\;O_2\;\rightarrow\;2\;ZnO\;+\;2\;SO_2[/tex]
Thus, 2 moles of ZnS will give 2 moles of ZnO.
The moles of 46.5 g ZnS:
Moles = [tex]\rm \dfrac{weight}{molecular\;weight}[/tex]
ZnS = [tex]\rm \dfrac{46.5}{97.474}[/tex]
ZnS = 0.477 mol
The moles of 13.3 grams Oxygen:
Oxygen = [tex]\rm \dfrac{13.3}{32}[/tex]
Oxygen = 0.415 mol.
The moles of 18.14 grams ZnO:
ZnO = [tex]\rm \dfrac{18.14}{81.38}[/tex]
ZnO = 0.222 mol
According to the balanced chemical equation,
2 moles ZnO = 2 moles ZnS
0.222 mol ZnO = 0.222 mol ZnS
2 moles ZnO = 3 moles Oxygen
0.222 mol ZnO = 0.334 mol of Oxygen
Since, both the reactants are in enough concentration,
The theoretical yield can be calculated with any of the reactants. The theoretical yield of ZnO from 46.5 grams of ZnS can be given as:
1 mole ZnS = 1 mol ZnO
0.477 mol of ZnS = 0.477 mol of ZnO.
The mass of 0.477 mol of ZnO:
Mass = moles [tex]\times[/tex] molecular weight
Mass of ZnO = 0.477 [tex]\times[/tex] 81.38
Mass of ZnO = 38.818 grams.
The theoretical yield of ZnO = 38.818 g.
The actual yield of ZnO = 18.14 g.
Percentage yield = [tex]\rm \dfrac{18.14}{38.81}\;\times\;100[/tex]
Percentage yield = 46.74 %
The percentage yield of the reaction has been 46.74%.
For more information about the percentage yield, refer to the link:
https://brainly.com/question/11715808
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1 point
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The atom of the reactants unbound , rearrange, and then rebond to form the product
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