Use this reaction to answer the question: 2KClO3 -> 2KCl + 3O2.

If I start with 65g KClO3, how many mol O2 will be formed? How many g O2?

If I need to produce 299g KCl, how many mol of KClO3 do i need to start with? How many g KClO3?

To determine the number of moles of O2 formed, we need to use the stoichiometry of the reaction.

From the balanced equation: 2KClO3 -> 2KCl + 3O2
We can see that 2 moles of KClO3 will produce 3 moles of O2.

Given that we start with 65g of KClO3, we can use the molar mass of KClO3 to convert grams to moles:

Molar mass of KClO3 = (1×39.1) + (1×35.45) + (3×16.00) = 122.55 g/mol

Number of moles of KClO3 = mass / molar mass = 65g / 122.55 g/mol ≈ 0.53 mol

Using the stoichiometry ratio, we can determine the number of moles of O2 formed:

Number of moles of O2 = Number of moles of KClO3 × (3 moles of O2 / 2 moles of KClO3)
= 0.53 mol × (3/2)
= 0.795 mol

To calculate the mass of O2 formed, we can use the molar mass of O2:

Molar mass of O2 = 2×16.00 = 32.00 g/mol

Mass of O2 = Number of moles of O2 × Molar mass of O2
= 0.795 mol × 32.00 g/mol
= 25.44 g

So, if you start with 65g of KClO3, you will produce 0.795 moles (or approximately 0.80 mol) of O2, which is equal to 25.44 grams of O2.

Now, let's calculate the amount of KClO3 needed to produce 299g of KCl:

Again, referencing the balanced equation: 2KClO3 -> 2KCl + 3O2

From the equation, we can see that 2 moles of KClO3 produce 2 moles of KCl.

Given that we want to produce 299g of KCl, we can use the molar mass of KCl to convert grams to moles:

Molar mass of KCl = (1×39.1) + (1×35.45) = 74.55 g/mol

Number of moles of KCl = mass / molar mass = 299g / 74.55 g/mol ≈ 4.01 mol

Since the stoichiometry ratio between KClO3 and KCl is 2:2, we need the same number of moles of KClO3 as KCl:

Number of moles of KClO3 = 4.01 mol

To determine the mass of KClO3 needed, we can use its molar mass:

Molar mass of KClO3 = (1×39.1) + (1×35.45) + (3×16.00) = 122.55 g/mol

Mass of KClO3 = Number of moles of KClO3 × Molar mass of KClO3
= 4.01 mol × 122.55 g/mol
= 490.96 g

So, you would need to start with approximately 491g of KClO3 to produce 299g of KCl.

To answer the first question using the given reaction: 2KClO3 -> 2KCl + 3O2, we can use stoichiometry.

Step 1: Convert grams of KClO3 to moles of KClO3.
To do this, you need to calculate the molar mass of KClO3 (potassium chlorate). The molar mass of K (potassium) is 39.1 g/mol, Cl (chlorine) is 35.5 g/mol, and O (oxygen) is 16.0 g/mol. Adding them up, the molar mass of KClO3 is 39.1 + 35.5 + (16.0 x 3) = 122.5 g/mol.

To convert grams to moles, divide the given mass (65g) by the molar mass of KClO3:
65g KClO3 * (1 mol/122.5g KClO3) = 0.531 mol KClO3

Step 2: Use the stoichiometric ratio to find moles of O2 produced.
According to the balanced equation, for every 2 moles of KClO3 consumed, 3 moles of O2 are produced. So, using the mole ratio:
0.531 mol KClO3 * (3 mol O2/2 mol KClO3) = 0.797 mol O2

Therefore, you will produce 0.797 moles of O2.

Step 3: Convert moles of O2 to grams of O2.
To do this, you need to calculate the molar mass of O2 (oxygen gas). The molar mass of O2 is 16.0 x 2 = 32.0 g/mol.

To convert moles to grams, multiply the moles of O2 (0.797 mol) by the molar mass of O2:
0.797 mol O2 * (32.0g O2/1 mol O2) = 25.5 g O2

Therefore, you will produce 25.5 grams of O2.

Now, let's move on to the second question.

If you need to produce 299g of KCl (potassium chloride), you can determine the amount of KClO3 (potassium chlorate) needed using stoichiometry again.

Step 1: Convert grams of KCl to moles of KCl.
To do this, you need to calculate the molar mass of KCl. The molar mass of K (potassium) is 39.1 g/mol, and Cl (chlorine) is 35.5 g/mol. Adding them up, the molar mass of KCl is 39.1 + 35.5 = 74.6 g/mol.

To convert grams to moles, divide the given mass (299g) by the molar mass of KCl:
299g KCl * (1 mol/74.6g KCl) = 4.01 mol KCl

Step 2: Use the stoichiometric ratio to find moles of KClO3 needed.
According to the balanced equation, for every 2 moles of KClO3 consumed, 2 moles of KCl are produced. So, using the mole ratio:
4.01 mol KCl * (2 mol KClO3/2 mol KCl) = 4.01 mol KClO3

Therefore, you will need 4.01 moles of KClO3.

Step 3: Convert moles of KClO3 to grams of KClO3.
To do this, multiply the moles of KClO3 (4.01 mol) by the molar mass of KClO3 (previously calculated as 122.5 g/mol):
4.01 mol KClO3 * (122.5g KClO3/1 mol KClO3) = 491 g KClO3

Therefore, you will need 491 grams of KClO3.