Consider the following reaction.
2KClO3(s) → 2KCl(s) + 3O2(g)
a. How many grams of KCl are produced from the complete reaction of 2.16 mol of KClO3?
161g
b. How many grams of KCl are produced from the complete reaction of 52.7 g of KClO3?
32.1 g
I agree.
To find the answers to these questions, we need to use stoichiometry and convert between moles and grams.
a. To find how many grams of KCl are produced from the complete reaction of 2.16 mol of KClO3, we can use the stoichiometric ratio between KClO3 and KCl in the balanced chemical equation.
From the balanced equation, we can see that 2 moles of KClO3 react to produce 2 moles of KCl. This means that the mole ratio between KClO3 and KCl is 2:2, which can be simplified to 1:1.
First, we need to convert the given moles of KClO3 to grams. To do this, we need to know the molar mass of KClO3, which can be calculated by adding up the atomic masses of the elements:
K: 39.10 g/mol
Cl: 35.45 g/mol
O: 16.00 g/mol (x3 since there are 3 oxygen atoms in KClO3)
Molar mass of KClO3 = 39.10 + 35.45 + (16.00 x 3) = 122.55 g/mol
Now, we can calculate the grams of KCl produced using the conversion factor:
2.16 mol KClO3 * (2 mol KCl / 2 mol KClO3) * (74.55 g KCl / 1 mol KCl) = 161 g KCl
Therefore, 2.16 mol of KClO3 produces 161 g of KCl.
b. To find how many grams of KCl are produced from 52.7 g of KClO3, we follow a similar process.
First, we need to convert the given grams of KClO3 to moles using the molar mass of KClO3:
52.7 g KClO3 * (1 mol KClO3 / 122.55 g KClO3) = 0.43 mol KClO3
Now, we can calculate the grams of KCl produced using the stoichiometry:
0.43 mol KClO3 * (2 mol KCl / 2 mol KClO3) * (74.55 g KCl / 1 mol KCl) = 32.1 g KCl
Therefore, 52.7 g of KClO3 produces 32.1 g of KCl.