A 115.0-g sample of oxygen was produced by heating 400.0 g of potassium chlorate (2KClO3 mc019-1.jpg 2KCI + 3O2) What is the percent yield of oxygen in this chemical reaction? a. 69.63% b. 73.40% c. 90.82% d. 136.2%

we are given with a 115 gram sample of oxygen to produce 400 gram potassium chlorate. The equation is 2KClO3 to form 2KCI + 3O2. 400 gram sample of potassium chlorate can be converted to 100% conversion to 156.67 gram of oxygen. The percent yield is 73.40 percent

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Mergus Mergus · Answer 2 · 2019-11-26T06:14:54+01:00

Answer:

b. 73.40%

Explanation:

The formula for the calculation of moles is shown below:

[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]

For [tex]KClO_3[/tex] :-

Mass of [tex]KClO_3[/tex] = 400.0 g

Molar mass of [tex]KClO_3[/tex] = 122.55 g/mol

The formula for the calculation of moles is shown below:

[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]

Thus,

[tex]Moles= \frac{400.0\ g}{122.55\ g/mol}[/tex]

[tex]Moles\ of\ KClO_3= 3.264\ mol[/tex]

From the given reaction:-

[tex]2KClO_3\rightarrow 2KCl + 3O_2[/tex]

2 moles of [tex]KClO_3[/tex] on reaction forms 3 moles of oxygen gas

1 mole of [tex]KClO_3[/tex] on reaction forms 3/2 moles of oxygen gas

3.264 moles of [tex]KClO_3[/tex] on reaction forms 1.5*3.264 moles of oxygen gas

Moles of oxygen gas = 4.896 moles

Molar mass of oxygen gas = 32 g/mol

Mass of oxygen gas = Moles * Molar mass = 156.7 g

he expression for the calculation of the percentage yield for a chemical reaction is shown below as:-

[tex]\%\ yield =\frac {Experimental\ yield}{Theoretical\ yield}\times 100[/tex]

Theoretical yield = 156.7 g

Given, Experimental yield = 115.0 g

Applying the values in the above expression as:-

[tex]\%\ yield =\frac{115.0}{156.7}\times 100[/tex]

[tex]\%\ yield =73.40\ \%[/tex]