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 23.642 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?
Here is a worked example of the exact problem you have (although the numbers are not the same). Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
2.000 g
To calculate the mass of the potassium chlorate originally present in the crucible, we first need to write the equation for the reaction.
The decomposition of potassium chlorate can be represented by the following equation:
2 KClO3(s) → 2 KCl(s) + 3 O2(g)
According to the stoichiometry of the reaction, for every 2 moles of KClO3 decomposed, we get 3 moles of oxygen gas.
Next, we can calculate the change in mass of the crucible.
Mass before heating = 24.426 g
Mass after heating = 23.642 g
Change in mass = Mass before heating - Mass after heating
= 24.426 g - 23.642 g
= 0.784 g
Since the reaction produces 3 moles of oxygen gas for every 2 moles of KClO3 decomposed, we can calculate the molar mass of KClO3:
Molar mass of KClO3 = (39.1 g/mol) + (35.45 g/mol) + (3 * 16.00 g/mol)
= 122.55 g/mol
Now, we can calculate the moles of KClO3 decomposed using the formula:
Moles of KClO3 = Change in mass / Molar mass of KClO3
= 0.784 g / 122.55 g/mol
Finally, we can calculate the mass of KClO3 originally present in the crucible:
Mass of KClO3 = Moles of KClO3 * Molar mass of KClO3.
To write the equation for the reaction, let's start with the information given:
Initial mass of crucible + potassium chlorate: 24.426 g
Final mass of crucible + any remaining substances: 23.642 g
The decrease in mass can be attributed to the decomposition of potassium chlorate into other substances.
The equation for the reaction can be written as follows:
2KClO3(s) → 2KCl(s) + 3O2(g)
In this reaction, potassium chlorate decomposes into potassium chloride and oxygen gas.
To calculate the mass of the potassium chlorate originally present in the crucible, we need to find the mass lost during the decomposition. The mass lost is equal to the initial mass minus the final mass:
Mass lost = Initial mass - Final mass
= 24.426 g - 23.642 g
= 0.784 g
This 0.784 g represents the mass of the oxygen that was produced during the decomposition. Since 2 moles of KClO3 decompose to yield 3 moles of O2, we can use the molar mass of oxygen (32 g/mol) to calculate the amount of potassium chlorate originally present:
1 mole of KClO3 → 1.5 moles of O2
X moles of KClO3 → 0.784 g of O2
Using molar ratios:
X = (0.784 g O2) / (32 g/mol)
X ≈ 0.0245 mol KClO3
Finally, we can calculate the mass of potassium chlorate originally present using its molar mass:
Mass of KClO3 = Moles of KClO3 × Molar mass of KClO3
= 0.0245 mol × (39.1 g/mol + 35.5 g/mol + 3(16.0 g/mol))
≈ 1.590 g
Therefore, the mass of potassium chlorate originally present in the crucible is approximately 1.590 g.