Question

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%

Answer 1

b. 73.40%

Step-by-step explanation:

The formula for the calculation of moles is shown below:

`moles = (Mass\ taken)/(Molar\ mass)`

For
`KClO_3` :-

Mass of
`KClO_3` = 400.0 g

Molar mass of
`KClO_3` = 122.55 g/mol

The formula for the calculation of moles is shown below:

`moles = (Mass\ taken)/(Molar\ mass)`

Thus,

`Moles= (400.0\ g)/(122.55\ g/mol)`

`Moles\ of\ KClO_3= 3.264\ mol`

From the given reaction:-

`2KClO_3\rightarrow 2KCl + 3O_2`

2 moles of
`KClO_3` on reaction forms 3 moles of oxygen gas

1 mole of
`KClO_3` on reaction forms 3/2 moles of oxygen gas

3.264 moles of
`KClO_3` 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:-

`\%\ yield =\frac {Experimental\ yield}{Theoretical\ yield}* 100`

Theoretical yield = 156.7 g

Given, Experimental yield = 115.0 g

Applying the values in the above expression as:-

`\%\ yield =(115.0)/(156.7)* 100`

`\%\ yield =73.40\ \%`

Answer 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