Question

The reaction below was performed with 0.342 grams of K2CO3 (molar mass = 138.21 g/mol). After the reaction was complete 0.0675 grams of CO2 was recovered. What was the percent yield to 2 significant figures?

K2CO3(s) --> K2O(s) + CO2(g)

The correct answer is 62 %, but I don't understand how they got there. Please help explain the steps

Answer 1

Following is the - complete Answer -&- Explanation: for the given Question, in......typed format...

Answer:

Percent yield of the reaction = 62.0 % ( i.e. correct up to 2 significant figures )

Explanation:

Following is the complete Explanation, for the above Answer...

• Given:

1. balanced reaction:  K₂CO₃ (s) K₂O (s) + CO₂ (g) --------------------- Equation - 1
2. Initial mass of potassium carbonate ( K₂CO₃) =  0.342 g ( grams )
3. Mass of carbon di-oxide ( CO₂) recovered, from reaction = 0.0675 g ( grams )
4. Molar mass of potassium carbonate ( K₂CO₃) = 138.21 g/mol  ( grams per mole )

• Step - 1:

​​​​​​​We have the following balanced reaction: i.e. the following:

K₂CO₃ (s) K₂O (s) + CO₂ (g) --------------------- Equation - 1

And, from the above balanced reaction, we are getting the following molar ratio:

molar ratio:   K₂CO₃ (s) : CO₂ (g) = 1 : 1

• Step - 2:

​​​​​​​Using the given information, we get the following:

Number of moles, of  K₂CO₃ = ( mass ) / ( molar mass ) = ( 0.342 g ) / ( 138.21 g/mol ) = 0.00247 moles ( approx. )

Therefore, according to the Equation - 1, given above, we will get the following:

Number of moles, of carbon di-oxide ( CO₂), obtained after complete reaction =  0.00247 moles

• Step - 3:

​​​​​​​We know the following:

Molar mass of CO₂ =  44.01 g/mol (approx. )

Mass of CO₂ (g) recovered from the reaction = ( 0.00247 mol ) x ( 44.01 g/mol ) = 0.1087 g ( grams ), approx.

Therefore, we can state the following:

1. Theoretical yield of the reaction = 0.1087 g ( grams ) of CO₂ (g)
2. Actual yield of the reaction = 0.0675 g ( grams ) of CO₂ (g), approx. ( i.e. given.... )

• ​​​​​​​Step - 4:

​​​​​​​Therefore, the percent yield, of the reaction will be as the following:

percent yield =  [ ( actual yield ) / ( theoretical yield ) ] x 100% = [ ( 0.0675 g ) / (0.1087 g ) ] x 100% = 62.097 %

percent yield = 62.097 %

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