15.0g of potassium trioxocarbonate(v) was crushed as heated with 0.1g of manganese (iv)oxide.
1) write the equation for the reaction.
2) calculate the mass of components produced assuming the reaction was complete.
3) find the volume of oxygen produced at 0.93×10^5nm^-2 pressure and at 23° by potassium chlorate(v)?
3kclo3-2kcl +3o2
Clarify please. Is that 15.0 g K2CO3 or 15.0 g KClO3?
Kclo3
kclo3
Question
1) To write the equation for the reaction, we need to understand the chemical formulas and charges of the compounds involved.
Potassium trioxocarbonate (v), also known as potassium carbonate, has the chemical formula K2CO3. Manganese (IV) oxide is represented by the formula MnO2.
The equation for the reaction can be written as follows:
2K2CO3 + MnO2 → 2K2O + MnCO3 + CO2
2) To calculate the mass of components produced, we need to determine the molar ratios between the reactants and products. From the balanced equation, we can see that 2 moles of potassium trioxocarbonate (v) react with 1 mole of manganese (IV) oxide to produce 2 moles of potassium oxide, 1 mole of manganese carbonate, and 1 mole of carbon dioxide.
Given that the mass of potassium trioxocarbonate (v) is 15.0g and the mass of manganese (IV) oxide is 0.1g, we can use the molar masses of the compounds to calculate the number of moles for each reactant.
The molar mass of potassium trioxocarbonate (v) (K2CO3) is calculated as follows:
K (39.1 g/mol) * 2 + C (12.0 g/mol) + 3 * O (16.0 g/mol) = 138.2 g/mol
Now, we can calculate the number of moles of potassium trioxocarbonate (v):
Number of moles = mass / molar mass = 15.0 g / 138.2 g/mol ≈ 0.108 mol
Similarly, we calculate the number of moles of manganese (IV) oxide:
Number of moles = mass / molar mass = 0.1 g / 86.9 g/mol ≈ 0.00115 mol
Since the balanced equation shows a 2:1 ratio between potassium trioxocarbonate (v) and manganese (IV) oxide, the number of moles of manganese (IV) oxide will be half of that of potassium trioxocarbonate (v) in order to react completely.
Thus, the theoretical number of moles of manganese (IV) oxide is 0.108 mol / 2 = 0.054 mol.
Using the stoichiometry of the balanced equation, we can determine the number of moles of products:
Potassium oxide (K2O): 2 * 0.054 mol = 0.108 mol
Manganese carbonate (MnCO3): 0.054 mol
Carbon dioxide (CO2): 0.054 mol
We can now calculate the mass of each component using the respective molar masses:
Mass of potassium oxide = number of moles * molar mass = 0.108 mol * (39.1 g/mol) = 4.2268 g
Mass of manganese carbonate = number of moles * molar mass = 0.054 mol * (114.9 g/mol) = 6.2046 g
Mass of carbon dioxide = number of moles * molar mass = 0.054 mol * (44.0 g/mol) = 2.376 g
Therefore, assuming the reaction goes to completion, the masses of the components produced are:
- Potassium oxide: 4.2268 g
- Manganese carbonate: 6.2046 g
- Carbon dioxide: 2.376 g
3) To find the volume of oxygen produced, we need to know the stoichiometry of the reaction involving potassium chlorate (KClO3). However, the original problem did not mention potassium chlorate.
If you provide the balanced equation for the reaction involving potassium chlorate and additional information such as the number of moles or mass of potassium chlorate, we can help you calculate the volume of oxygen produced using the ideal gas law (PV = nRT).