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Answer each of the following questions completelby Type your work, do not handwrite Use complete sentences with proper prammar and spelling for credit. Number (and letter, If applicable) your answers clearly. Please make them easy to find. EXPERIMENT PRE-LAB QUESTIO While performing Experiment 6, after two successive heatines a student collected the following data: Mass of crucible and o Mass of HCIO Mass of crucible/ld and residue 22.11eating Mass of cruobic/lid and residue 22.09 (2 heatrg) 1.25 Using the information shown above, answer problems 1-6 showing all caloulations and work for credit. 1 Wll this student have to do a third heating? Include a calouation as proot of your answer 2 Cakulate the mass of the residue left behind in the crucble after the second heating. 3 Calculate the mass kost during heating 4 calculate the experimental % cogerin the potassium criorate sore 5 Calouste the theoretical % appen n potassium chlorate. 6 Calculate the % error in the experimental % oaygen determination. 7 Write the balanced equation for the decomposition af potassium chlorate; include the states of al species irn 8 (a) After each heating, reection whry shoukd you wait unti your crucibie is room temperature before weighing t? (b) If you weigh a warm crucible, would this be considered a systematic or random error? 9 Potassium chiorate is a strong ouidiaing agent lke the sodikam chlorate used in Experiment 2. How should the student dispose of excess potassium chiorate? 10 Potassium chiorate and potassium chionde kock very simlar but react very differently with siver nitrate. (a) Wribe the balanced equstions for the reaction of silver nitrate with potassium chlorate and the reaction of s nitrate with potassium chloride. Indicate the states of all species using a solublity table. (b) How might a student verify that the residue left in his crucbie at the end of Experiment 6 has changed from potassium chiorate to potassium chloride? Cuyamace College Experiment 6 Araysis of Potasskum Obtste
Explanation / Answer
1. mass of KClO3 = 1.25 g
mass of O2 formed = 1.25 g x 32 g/mol x 3/122.55 g/mol x 2
= 0.50 g
Final weight of KCl + crucible + lid = 21.32 + (1.25 - 0.50)
= 22.07 g
Since the final weight obtained was 22.09 g which is greater than the required weight 22.07 g, we do need a third heating.
2. mass of residue left after second heating = 22.09 - 21.32 = 0.77 g
3. mass lost after second heating = 0.48 g
4. experimental %O2 in the sample = 0.48 x 100/1.25 = 38.4%
5. Theoretical %O2 in the sample = 0.50 x 100/1.25 = 40.0%
6. %error = (40 - 38.4) x 100/40 = 4.0%
7. Equation,
2KClO3(s) --> 2KCl(s) + 3O2(g)
8. (a) after heating the hot sample takes up moisture from the system and this causes higher mass than expected and thus we should wait it to cool down.
(b) This causes a random error in the measurement.
9. We must add a neutralizing agent such as a sodium bisulfite a reducing agent prior to its disposal.
10. reaction
(a) KClO3 + AgNO3 --> AlClO3 (aq) + KNO3
KCl + AgNO3 --> AgCl (ppt) + KNO3
(b) As can be seen from above, KCl forms a white insoluble precipitate with AgNO3 of AgCl, whereas, KClO3 forms a soluble solid AgClO3. Thus the two salts can easily be distinguished.
