15.0g of potassium trioxochlorate (v) was crushed and heated with about 0.1g of manganese (iv) oxide
1.write an equation for the reaction
2.calculate the mass of c that would be produced assuming the reaction was complete
Molar mass of oxygen=02
I will find a perfect school as this
To answer these questions, we need to understand the chemical reaction between potassium trioxochlorate (V) and manganese (IV) oxide.
1. Equation for the reaction:
The reaction between potassium trioxochlorate (V) and manganese (IV) oxide can be represented by the following chemical equation:
2 KClO3 + 3 MnO2 → 2 KCl + 3 MnO4
This equation shows that two moles of potassium trioxochlorate (V) react with three moles of manganese (IV) oxide to produce two moles of potassium chloride and three moles of manganese (IV) oxide.
2. Calculating the mass of carbon (C) produced:
To calculate the mass of carbon (C) produced, we need to determine the limiting reactant first. The limiting reactant is the one that gets completely consumed in the reaction, limiting the amount of products that can be formed.
To find the limiting reactant, we compare the number of moles of each reactant. Given that we have 15.0g of potassium trioxochlorate (V) and 0.1g of manganese (IV) oxide, we can convert these masses to moles.
The molar mass of potassium trioxochlorate (V), KClO3, is:
K = 39.1 g/mol
Cl = 35.5 g/mol
O = 16.0 g/mol (x3 since there are three oxygen atoms)
Molar mass of KClO3 = 39.1 + 35.5 + (16.0 x 3) = 122.5 g/mol
Number of moles of KClO3 = Mass / Molar mass = 15.0 g / 122.5 g/mol
Similarly, the molar mass of manganese (IV) oxide, MnO2, is:
Mn = 54.9 g/mol
O = 16.0 g/mol (x2 since there are two oxygen atoms)
Molar mass of MnO2 = 54.9 + (16.0 x 2) = 86.9 g/mol
Number of moles of MnO2 = Mass / Molar mass = 0.1 g / 86.9 g/mol
We compare the moles of each reactant and find that the moles of MnO2 are significantly lower than the moles of KClO3. Therefore, MnO2 is the limiting reactant.
From the balanced equation, we know that 2 moles of KClO3 produce 1 mole of carbon (C). So, from the moles of MnO2, we can calculate the moles of carbon (C) produced:
Moles of C = Moles of MnO2 x (1/2)
Finally, to calculate the mass of carbon (C) produced, we multiply the number of moles by the molar mass of carbon (C) (assuming complete reaction):
Mass of C = Moles of C x Molar mass of C
Please provide the molar mass of carbon (C) to complete the calculation.
2KClO3 ==> 2KCl + 3O2
You must have made a typo. There is no c. You must have meant how much O2 is evolved.
mols KClO3 = grams/molar mass = 15/122.5 = 0.122
Convert to mols O2 produced. 0.122 mols KClO3 x (3 mols O2/2 mols KClO3) = 0.183
Convert to mass O2. grams O2 = mols O2 x molar mass O2 = ?
Post your work if you get stuck.