Question

Lab Questions

Part 1

Mass of test tube + MnO2 = 31.562 g

Mass of test tube + MnO2 + KClO3 = 32.489 g

Mass of test tube after heating = 32.173 g

1. Based on the data above, what is the mass of O2 released in grams?

2. What’s is the experimental yield of O2 in moles?

3. What is the theoretical yield of O2 in moles?

4. Calculate % error

% error = |(theoretical-experimental) / (theoretical) | x 100%

Part 2

Mass of test tube + MnO2 = 31.209 g

Mass of test tube + MnO2 + unknown = 32.415g

Mass of test tube after heating = 32.273 g

1. With the information in part 2, mass of unknown in grams?

2. What is the mass of O2 released in grams?

3. How many moles of O2 released?

4. How many moles of KClO3 in mixture?

5. What is the mass of KClO3 in mixture in grams?

6. What is the mass percent KClO3 in unknown mixture sample?

Answer 1

Sol.

Part 1 :

As Decomposition Reaction of KClO3 in presence of MnO2 :

2KClO3 ----> 2KCl + 3O2  

(1)

As Mass of O2 released in grams

= Mass of test tube after heating - Mass of test tube + MnO2

= 32.173 g - 31.562 g

= 0.611 g

(2)

As Molar Mass of O2 = 32 g/mol

So , Experimental yield of O2 in moles  

= 0.611 g / 32 g/mol

=   0.01909 mol  

(3)

As Mass of KClO3 = Mass of test tube + MnO2 + KClO3 - Mass of test tube + MnO2  

= 32.489 g - 31.562 g

= 0.927 g  

Molar Mass of KClO3 = 122.55 g/mol

So , Moles of KClO3 = 0.927 g / 122.55 g/mol

= 0.00756 mol  

From reaction ,  

Moles of O2 = Moles of KClO3 × 3 / 2  

= 0.00756 mol × 3 / 2 = 0.01134 mol

Therefore , Theoretical yield of O2 in moles

=   0.01134 mol  

(4)

Percentage Error

= | ( Theoretical yield of O2 - Experimental  yield of O2 ) / Theoretical yield of O2 | × 100

= | ( 0.01134 mol - 0.01909 mol ) / 0.01134 mol | × 100

= | - 0.00775 mol / 0.01134 mol | × 100

= 0.683421  × 100

=    68.34 %