If 100 grams of KClO3 are heated to form KCl and O2 according to the equation
KClO3 �¨ KCl + O2 (Hint: You need to balance the equation)
a. How many moles of KClO3 reacted?
b. How many moles of KCl are formed?
c. How many grams of KCl are formed?
d. how many moles of O2 are formed?
e. How many grams of O2 are formed?
2KClO3 ==> 2KCl + 3O2
a. How many moles of KClO3 reacted?
mols KClO3 = grams/molar mass.
b. How many moles of KCl are formed?
mols KCl = mols KClO3 x (2 mol KCl/2 mol KClO3) = mols KClO3 x 2/2 = ?
c. How many grams of KCl are formed?
grams KCl = mols KCl x molar mass KCl
d. how many moles of O2 are formed?
mols O2 = mols KClO3 x (3 mols O2/2 mols KClO3) = mols KClO3 x 3/2 = ?
e. How many grams of O2 are formed?
gO2 = mols O2 x molar mass O2.
LADUU
To answer these questions, let's first balance the equation:
2 KClO3 → 2 KCl + 3 O2
Now we can proceed to answer the questions:
a. To find the number of moles of KClO3 reacted, we need to use the molar mass of KClO3, which is 122.55 g/mol.
Moles of KClO3 = Mass of KClO3 / Molar mass of KClO3
= 100 g / 122.55 g/mol
≈ 0.8166 mol
b. From the balanced equation, we can see that for every 2 moles of KClO3 reacted, 2 moles of KCl are formed.
Moles of KCl = Moles of KClO3
= 0.8166 mol
c. To find the mass of KCl formed, we need to use the molar mass of KCl, which is 74.55 g/mol.
Mass of KCl = Moles of KCl * Molar mass of KCl
= 0.8166 mol * 74.55 g/mol
≈ 60.82 g
d. From the balanced equation, we can see that for every 2 moles of KClO3 reacted, 3 moles of O2 are formed.
Moles of O2 = (Moles of KClO3 / 2) * 3
= (0.8166 mol / 2) * 3
≈ 1.225 mol
e. To find the mass of O2 formed, we need to use the molar mass of O2, which is 32.00 g/mol.
Mass of O2 = Moles of O2 * Molar mass of O2
= 1.225 mol * 32.00 g/mol
≈ 39.2 g
So, the answers to the questions are:
a. Approximately 0.8166 moles of KClO3 reacted.
b. Approximately 0.8166 moles of KCl are formed.
c. Approximately 60.82 grams of KCl are formed.
d. Approximately 1.225 moles of O2 are formed.
e. Approximately 39.2 grams of O2 are formed.
To solve this problem, we need to follow a step-by-step process. First, we need to balance the equation:
2KClO3 → 2KCl + 3O2
Once we have a balanced equation, we can use the stoichiometry of the reaction to answer the questions. The stoichiometry tells us the molar ratios between the reactants and the products.
a. To find the number of moles of KClO3 reacted, we need to know the molar mass of KClO3. The molar mass of KClO3 can be calculated by adding the molar masses of potassium (K), chlorine (Cl), and oxygen (O).
K: 39.1 g/mol
Cl: 35.5 g/mol
O: 16.0 g/mol
Molar mass of KClO3: 39.1 + 35.5 + (3 * 16.0) = 122.55 g/mol
Now we can calculate the moles of KClO3:
Moles of KClO3 = Mass of KClO3 / Molar mass of KClO3
= 100 g / 122.55 g/mol
≈ 0.816 mol
b. According to the balanced equation, the molar ratio between KClO3 and KCl is 2:2. Therefore, the number of moles of KCl formed is the same as the number of moles of KClO3 reacted, which is 0.816 mol.
c. To find the grams of KCl formed, we need to know the molar mass of KCl:
K: 39.1 g/mol
Cl: 35.5 g/mol
Molar mass of KCl: 39.1 + 35.5 = 74.6 g/mol
Now we can calculate the grams of KCl:
Grams of KCl = Moles of KCl * Molar mass of KCl
= 0.816 mol * 74.6 g/mol
≈ 60.89 g
d. According to the balanced equation, the molar ratio between KClO3 and O2 is 2:3. Therefore, the number of moles of O2 formed can be calculated by multiplying the moles of KClO3 reacted by the molar ratio:
Moles of O2 = Moles of KClO3 * (3 mol O2 / 2 mol KClO3)
= 0.816 mol * (3/2)
≈ 1.224 mol
e. To find the grams of O2 formed, we need to know the molar mass of O2:
O: 16.0 g/mol
Molar mass of O2: 2 * 16.0 = 32.0 g/mol
Now we can calculate the grams of O2:
Grams of O2 = Moles of O2 * Molar mass of O2
= 1.224 mol * 32.0 g/mol
≈ 39.17 g
So, the answers to the questions are:
a. 0.816 moles of KClO3 reacted
b. 0.816 moles of KCl are formed
c. 60.89 grams of KCl are formed
d. 1.224 moles of O2 are formed
e. 39.17 grams of O2 are formed.