15.0g of potassium trioxochlorate(v) was crushed and heated with about 0.1g of manganese(iv)oxide, calculate the mass of chloride that would be produced assuming the reaction was complete
For whatever it's worth, the correct IUPAC name for KClO3 is potassium chlorate.
..................2KClO3 ==> 2KCl + 3O2
mols KClO3 = g/molar mass = 15.0/122.5 = 0.122
2 mols KClO3 will produce 2 mols KCl; therefore, 0.122 mols KClO3 will produce 0.122 mols KCl. Convert to g. gKCl = mols KCl x molar mass KCl = ?
Well, well, well, looks like we have a chemical reaction going on here! Let's do some clown math, shall we?
So, with potassium trioxochlorate(V) (KClO3), we have 3 oxygen atoms for each chlorine atom. And since there is 15.0g of KClO3, we need to figure out how much chlorine (Cl) is in there.
The molar mass of KClO3 is about 122.55 g/mol. And since there is only one chlorine in KClO3, the molar mass of chlorine is about 35.45 g/mol.
Using the molar ratio, we can convert grams of KClO3 to grams of chlorine, like this:
(15.0g KClO3) x (1 mol KClO3 / 122.55 g KClO3) x (1 mol Cl / 1 mol KClO3) x (35.45 g Cl / 1 mol Cl) = approximately 4.33g Cl
So, we have about 4.33g of chlorine.
Now, since 0.1g of manganese(IV) oxide (MnO2) is present, and it's an oxidizing agent, we can assume it will react with all of the chlorine, and form manganese(IV) chloride (MnCl2). The molar mass of MnO2 is about 86.93 g/mol, and MnCl2 is about 125.84 g/mol.
Using the molar ratio, we can convert grams of MnO2 to grams of MnCl2, like this:
(0.1g MnO2) x (1 mol MnO2 / 86.93 g MnO2) x (1 mol MnCl2 / 1 mol MnO2) x (125.84 g MnCl2 / 1 mol MnCl2) = approximately 0.143g MnCl2
So, approximately 0.143g of manganese(IV) chloride is produced.
Now, since the reaction is balanced, for each chlorine atom, we get one chloride (Cl-) ion.
Therefore, the mass of chloride produced would also be approximately 0.143g.
Hope that didn't cause any chemical confusion! Feel free to ask if you need more clown-tastic calculations!
To determine the mass of chloride produced, we need to balance the chemical equation for the reaction between potassium trioxochlorate(V) and manganese(IV) oxide.
The balanced chemical equation for this reaction is as follows:
2 KClO3 + MnO2 → 2 KCl + MnO4
From the balanced equation, we can see that 2 moles of potassium trioxochlorate(V) (KClO3) react with 1 mole of manganese(IV) oxide (MnO2) to produce 2 moles of potassium chloride (KCl).
Now, let's calculate the number of moles of potassium trioxochlorate(V) and manganese(IV) oxide based on the given masses.
Molar mass of KClO3:
K: 39.10 g/mol
Cl: 35.45 g/mol
O: 16.00 g/mol
Total: 39.10 + 3(35.45) + 3(16.00) = 122.55 g/mol
Number of moles of KClO3 = mass / molar mass
Number of moles of KClO3 = 15.0 g / 122.55 g/mol = 0.1225 mol
Number of moles of MnO2 = mass / molar mass
Number of moles of MnO2 = 0.1 g / 86.94 g/mol (molar mass of MnO2) = 0.00115 mol
Now, we can use the balanced equation to determine the moles of KCl produced.
From the balanced equation, we can see that 2 moles of KClO3 produce 2 moles of KCl.
Since the stoichiometric ratio is 2:2 (KClO3:KCl), the number of moles of KCl produced is equal to the number of moles of KClO3 used.
Therefore, the number of moles of KCl produced = 0.1225 mol
Finally, we can calculate the mass of KCl produced.
Molar mass of KCl:
K: 39.10 g/mol
Cl: 35.45 g/mol
Total: 39.10 + 35.45 = 74.55 g/mol
Mass of KCl produced = number of moles of KCl × molar mass of KCl
Mass of KCl produced = 0.1225 mol × 74.55 g/mol = 9.13 g
Therefore, the mass of chloride (KCl) produced is approximately 9.13 grams.
To calculate the mass of chloride that would be produced in the reaction, we need to determine the limiting reactant first. The limiting reactant is the reactant that is completely consumed and limits the amount of product that can be formed.
Let's start by writing the balanced chemical equation for the reaction between potassium trioxochlorate(v) (KClO3) and manganese(iv) oxide (MnO2):
2 KClO3 + MnO2 → 2 KCl + 3 O2 + MnO2
From the balanced equation, we can see that 2 moles of KClO3 react with 1 mole of MnO2 to produce 2 moles of KCl. Therefore, the mole ratio between KClO3 and KCl is 2:2, which simplifies to 1:1.
Next, we need to convert the given masses of KClO3 and MnO2 into moles using their respective molar masses:
- Molar mass of KClO3 = (39.1 g/mol) + (35.5 g/mol) + (3 * 16.0 g/mol) = 122.6 g/mol
- Moles of KClO3 = 15.0 g / 122.6 g/mol = 0.1223 mol
- Molar mass of MnO2 = (55.8 g/mol) + (2 * 16.0 g/mol) = 87.8 g/mol
- Moles of MnO2 = 0.1 g / 87.8 g/mol = 0.0011 mol
Now, we compare the mole ratio between KClO3 and KCl with the actual moles of KClO3 and MnO2 to determine the limiting reactant.
Since the mole ratio is 1:1, and we have 0.1223 moles of KClO3, we can conclude that only 0.1223 moles of KCl can be produced.
Now, using the molar mass of KCl (39.1 g/mol + 35.5 g/mol = 74.6 g/mol), we can calculate the mass of KCl:
Mass of KCl = 0.1223 mol * 74.6 g/mol = 9.13 g
Therefore, assuming the reaction is complete, 9.13 g of chloride (KCl) would be produced.