if 50.0 g of KCl reacts with 50.0 g of O2 to produce KClO3 according to the equation, how many grams of KClO3 will be formed?
Well, it seems like we have a salty situation here with some KCl and a breath of fresh air with some O2! According to the equation, let's find out how much KClO3 we'll have in the end.
The balanced equation tells us that 4 moles of KCl react with 3 moles of O2 to produce 2 moles of KClO3. To figure out the grams, we need to convert the grams of KCl and O2 into moles.
The molar mass of KCl is 74.6 g/mol, so 50.0 g would be approximately 0.67 moles of KCl. The molar mass of O2 is 32.0 g/mol, so 50.0 g would be approximately 1.56 moles of O2.
Now, let's look at the ratio from the balanced equation. If 4 moles of KCl react with 3 moles of O2 to produce 2 moles of KClO3, we can set up a proportion:
0.67 moles of KCl / 4 moles of KClO3 = 1.56 moles of O2 / x moles of KClO3
Solving for x, we get x ≈ 9.35 moles of KClO3. Now, we can convert this back into grams:
9.35 moles of KClO3 * (122.6 g/mol of KClO3) ≈ 1144.81 g of KClO3.
So, approximately 1144.81 grams of KClO3 will be formed. That's quite a potassium-rich explosion if you ask me!
The balanced equation for the reaction is:
4KCl + O2 -> 2KClO3
From the equation, we can see that 4 moles of potassium chloride (KCl) react with 1 mole of oxygen (O2) to produce 2 moles of potassium chlorate (KClO3).
Step 1: Convert grams to moles for KCl and O2.
To convert grams to moles, we need the molar mass of each substance. The molar mass of KCl is 74.55 g/mol and the molar mass of O2 is 32.00 g/mol.
Moles of KCl = mass of KCl / molar mass of KCl
= 50.0 g / 74.55 g/mol
= 0.6705 mol
Moles of O2 = mass of O2 / molar mass of O2
= 50.0 g / 32.00 g/mol
= 1.5625 mol
Step 2: Determine the limiting reactant.
The limiting reactant is the substance that is completely consumed in the reaction, limiting the amount of product that can be formed. To determine the limiting reactant, we need to compare the mole ratios from the balanced equation.
Mole ratio of KCl to KClO3 = 4 moles KCl : 2 moles KClO3 = 4:2 = 2:1
Mole ratio of O2 to KClO3 = 1 mole O2 : 2 moles KClO3 = 1:2
Since the mole ratio of O2 to KClO3 is smaller than that of KCl to KClO3, it means that O2 is the limiting reactant.
Step 3: Calculate the moles of KClO3 formed.
From the balanced equation, we know the mole ratio of O2 to KClO3 is 1:2. Therefore:
Moles of KClO3 = 2 * moles of O2
= 2 * 1.5625 mol
= 3.125 mol
Step 4: Convert moles of KClO3 to grams.
To convert moles to grams, we use the molar mass of KClO3, which is 122.55 g/mol.
Mass of KClO3 = moles of KClO3 * molar mass of KClO3
= 3.125 mol * 122.55 g/mol
= 382.969 g
Therefore, approximately 382.97 grams of KClO3 will be formed.
To find out how many grams of KClO3 will be formed, we need to determine the limiting reactant in the given reaction.
First, let's write and balance the chemical equation for the reaction:
2 KCl + 3 O2 -> 2 KClO3
The molar masses of the reactants and the product are as follows:
Molar mass of KCl = 39.10 g/mol
Molar mass of O2 = 32.00 g/mol
Molar mass of KClO3 = 122.55 g/mol
Now we need to calculate the number of moles for each reactant using the given masses:
Number of moles of KCl = 50.0 g / 39.10 g/mol = 1.28 mol (rounded to two decimal places)
Number of moles of O2 = 50.0 g / 32.00 g/mol = 1.56 mol (rounded to two decimal places)
According to the balanced equation, the stoichiometry of the reaction is 2 moles of KCl to 2 moles of KClO3 and 3 moles of O2 to 2 moles of KClO3. Therefore, KCl is the limiting reactant because it produces less moles of KClO3 than O2 does.
Now, let's calculate the number of moles of KClO3 formed based on the stoichiometry:
Number of moles of KClO3 = 1.28 mol KCl x (2 mol KClO3 / 2 mol KCl) = 1.28 mol (rounded to two decimal places)
Finally, we can calculate the mass of KClO3 formed using the molar mass:
Mass of KClO3 = 1.28 mol KClO3 x 122.55 g/mol = 156.90 g (rounded to two decimal places)
Therefore, 156.90 grams of KClO3 will be formed in the reaction.
2K Cl + 3O2 --> 2KClO3
KCl = 39+ 35.5 = 74.5 g/mol
O = 16 g/mol so O2 is 32 g/mol
I need 3 mols of O2 for every 2 mols of KCl
50 g KCl = 50/74.5 = .671 mols of KCl
so need .671 * 3/2 = 1.007 mols of O2 = 32 g
SO I have MORE O2 than I need, just use the .671 mols of KCl, limiting reagent
now
I get 1 mol KClO3 for every mol of KCl
so I get .671 mols of KClO3
74.5 + 3*16 = 122.5 grams/mol of KClO3
.671 * 122.5 = 2.2 grams