An unknown mixture containing KClO3 was added to an ignition tube weighting 41.960g, The tube and mixture was then found to weight 43.772g. After the first heating, the tube and its contents weighed 43.621g.
After the second and third heatings, it weighed 43.332g.
1)What mass of oxigen was liberated from the unknown mixture?
2)what mass of KClO3 was contained in the original mixture?
3)what was the percentage of KClO3 in the original mixture?
Note: weigh is spelled without the final t.
43.772 g = mass KClO3 mixture + tube
-41. 960 = mass tube
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?? = mass KClO3 initially.
43.772 = mass KClO3 mixtur + tube
-43.332 = mass after final heating
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?? = loss in mass = mass oxygen
%oxygen = (mass O2/mass KClO3)*100 =
is the lost in mass O? or H2O? sorry.
To solve these questions, we need to use the concept of conservation of mass. The mass of the reactant(s) should be equal to the mass of the product(s).
1) Mass of oxygen liberated:
To find the mass of oxygen liberated, we need to calculate the difference in mass before and after the heating. The initial mass is 43.772g, and after the first heating, it is 43.621g. The difference is:
Mass of oxygen liberated = Initial mass - Mass after 1st heating
Mass of oxygen liberated = 43.772g - 43.621g
Mass of oxygen liberated = 0.151g
Therefore, the mass of oxygen liberated from the unknown mixture is 0.151g.
2) Mass of KClO3:
To find the mass of KClO3 in the original mixture, we need to calculate the difference in mass between the initial mixture and the final mass after all the heatings. The initial mass is 41.960g, and the final mass after the third heating is 43.332g. The difference is:
Mass of KClO3 = Final mass - Initial mass
Mass of KClO3 = 43.332g - 41.960g
Mass of KClO3 = 1.372g
Therefore, the mass of KClO3 contained in the original mixture is 1.372g.
3) Percentage of KClO3 in the original mixture:
To find the percentage of KClO3, we divide the mass of KClO3 by the initial mass of the mixture and multiply by 100.
Percentage of KClO3 = (Mass of KClO3 / Initial mass) * 100
Percentage of KClO3 = (1.372g / 41.960g) * 100
Percentage of KClO3 ≈ 3.27%
Therefore, the percentage of KClO3 in the original mixture is approximately 3.27%.