A crucible containing potassium chlorate has a mass is 24.426 g. The crucible is heated according to the lab procedure for this week's experiment and is found to weigh 23.202 g. Write an equation for the reaction and, assuming that all the chlorate decomposed, calculate the mass of the potassium chlorate originally present in the crucible?
24.426g = mass xble + KClO3
23.202g = loss in mass upon heating
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1.224 g = mass O2 evolved.
2KClO3 ==> 2KCl + 3O2
mols O2 = grams/molar mass = 1.224/32 = about 0.038 (but you do it more accurately).
Convert to mols KClO3.
0.038 mol O2 x (2 mol KClO3/3 mol O2) = 0.038 x 2/3 = 0.026 mol KClO3.
g KClO3 = mols x molar mass = 0.026 x 122.6 = about 3.126 g = mass KClO3 originally present.
To determine the mass of potassium chlorate originally present in the crucible, we need to understand the reaction that took place and calculate the change in mass.
The equation for the decomposition of potassium chlorate is:
2KClO₃ → 2KCl + 3O₂
According to the equation, potassium chlorate decomposes into potassium chloride and oxygen gas. Now, let's calculate the change in mass.
Change in Mass = Final Mass - Initial Mass
Given:
Initial Mass (before heating) = 24.426 g
Final Mass (after heating) = 23.202 g
Change in Mass = 23.202 g - 24.426 g
Change in Mass = -1.224 g
Since the change in mass is negative, it indicates a loss of mass during the reaction. This loss of mass corresponds to the oxygen gas that was released during the decomposition of potassium chlorate.
Next, we need to calculate the mass of potassium chlorate originally present. Since we know the molar mass of potassium chlorate (KClO₃) is 122.55 g/mol, we can use this information to find the number of moles of potassium chlorate.
Number of Moles = Mass / Molar Mass
Number of Moles of KClO₃ = -1.224 g / 122.55 g/mol
Number of Moles of KClO₃ = -0.009975 mol
Since moles cannot be negative, we need to take the absolute value of the number of moles. Therefore, the number of moles of potassium chlorate originally present is approximately 0.009975 mol.
Finally, to calculate the mass of potassium chlorate originally present, we multiply the number of moles by the molar mass:
Mass of KClO₃ = Number of Moles * Molar Mass
Mass of KClO₃ = 0.009975 mol * 122.55 g/mol
Mass of KClO₃ = 1.2271 g
Therefore, the mass of potassium chlorate originally present in the crucible is approximately 1.2271 grams.