Ex 5

‫بسم ال الرحمن الرحيم‬ Name: ayedzahran Group: Day:

Experiment 5 "Decomposition of potassium chlorate" _________________________________________________________________ I. Observations and results: Part A Mass of test tube and MnO2 Mass of test tube, MnO2 and KClO3 Mass of test tube, MnO2 and KCl residue Mass of KClO3 added Mass of oxygen lost % oxygen in potassium chlorate (experimental) % oxygen in potassium chlorate (theoretical)

10 g 11 g 10.61 g 1g 0.39 g (1) 39 % (2) 39.18% (3)

Part B Mass of test tube and MnO2 Mass of test tube, MnO2 and mixture Mass of test tube, MnO2 and residue Mass of oxygen lost Moles of oxygen lost Moles of potassium chlorate in mixture % potassium chlorate in the mixture

14.58 g 15.58 g 15.5 g 0.08 (4) 5*10-3 mol (5) 1.67*10-3 mol (6) 20.4% (7)

II. Calculation: (1) Mass of oxygen = 11-10.61=0.39 g (2) % oxygen =(m(O2)/m(KClO3))*100%=)0.39/1)*100%=39% (3) From 1 mol KClO3 % oxygen =(48/122.5)*100%=39.18% (4) Mass of oxygen=15.58-15.5=0.08 g (5) n=0.08/16= 5*10-3 mol

(6) KClO3 : oxygen

1: 3 so n (KClO3) = 5*10-3 /3=1.67*10-3 mol (7) Mass KClO3 =1.67*10-3*122.5 =0.204 g

% potassium chlorate=(0.204/1)*100%=20.4% III. Answer the following questions: 1. How would your experiment results be effected if : a. Heating was not strong enough to decompose potassium chlorate completely The potassium chlorate will not decompose completely so the mass of residue will increase and mass of oxygen lost will decrease so %of oxygen will decrease b. some soot from the burner was deposited on the test tube The soot will effect on weighting which we take after remove the test tube from burner (this value will increase) then the mass of oxygen will decrease because the addition soot become over oxygen final %oxygen will decrease c. Manganese dioxide was not added Manganese dioxide is used as catalyst, and we know the catalyst not part of reaction it's faster the reaction also, so if we don’t add it the reaction become slower than if we add the catalyst, in the result the reaction will complete but in more time 2. If after heating 1.5 g of sodium bicarbonate – sodium carbonate mixture till

the complete decomposition of sodium bicarbonate the loss of the mass was 0.369 g, what is the percentage of sodium bicarbonate in the mixture? 2NaHCO3 Na2CO3+H2O+CO2 2NaHCO3(s)+ Na2CO3(s) Na2CO3(g)+ Na2CO3+ H2O+ CO2 mmix =m NaHCO3 +m Na2CO3=1.5 g ………..(1) m Na2CO3+m Na2CO3=1.05-0.369=1.131 g …………………..(2) From equation of reaction: n Na2CO3=(1/2)n NaHCO3 From (1) & (2) M NaHCO3+1.131-m Na2CO3 =1.5 M NaHCO3+1.131-(n Na2CO3 *M Na2CO3) = 1.5 M NaHCO3+1.131-((1/2)(n NaHCO3)(106)=1.5 M NaHCO3=1g % NaHCO3 = 1/1.5*100%=66.7% Not: For experiment number , For calculation number , For the solution of answers .