what mass of barium sulfate can be produced when 100 ml 0.100 M of solution of barium chloride is mixed with 100.0 ml of a 0.100 M solution of Iron (III) sulfate
Well, it seems like we have a chemical party going on here! Let's crunch some numbers and see what kind of fun we can have.
To find the mass of barium sulfate produced, we need to determine the limiting reagent. This is the reactant that will be completely consumed, thus determining the maximum amount of product that can be formed.
So, let's compare the moles of barium chloride and iron (III) sulfate in your solutions. We'll use the formula:
moles = concentration (M) × volume (L)
For barium chloride:
moles = 0.100 mol/L × 0.100 L = 0.010 mol
For iron (III) sulfate:
moles = 0.100 mol/L × 0.100 L = 0.010 mol
Since the moles of both reactants are equal, none of them is limiting. This means we have a balanced party with equal amounts of both guests!
Now, let's see what happens when these party animals react. The balanced chemical equation is:
3 BaCl2(aq) + Fe2(SO4)3(aq) -> Ba3(SO4)2(s) + 6 Cl-(aq) + 2 Fe3+(aq)
From the equation, we can see that 3 moles of barium chloride react with 1 mole of iron (III) sulfate to produce 1 mole of barium sulfate.
Since we have an equal amount of both reactants (0.010 mol), we can safely say that the maximum amount of barium sulfate that can be produced is 0.010 mol.
Now, to find the mass:
mass = moles × molar mass
The molar mass of barium sulfate (BaSO4) is approximately 233.38 g/mol.
mass = 0.010 mol × 233.38 g/mol = 2.33 g
So, the mass of barium sulfate that can be produced is approximately 2.33 grams.
And voila! The barium sulfate party is complete. Enjoy the chemical shenanigans!
To determine the mass of barium sulfate that can be produced, we need to first find the limiting reagent between barium chloride (BaCl2) and iron (III) sulfate (Fe2(SO4)3), as well as the balanced chemical equation.
The balanced chemical equation for the reaction between barium chloride and iron (III) sulfate is:
3 BaCl2 + Fe2(SO4)3 → 2 FeCl3 + 3 BaSO4
From the balanced equation, we can see that the molar ratio between BaCl2 and BaSO4 is 3:3, or 1:1. Therefore, the limiting reagent is the one that will yield the least amount of the desired product, which is BaCl2.
To find the limiting reagent, we need to calculate the number of moles of each reagent.
For barium chloride (BaCl2):
Molar mass of BaCl2 = atomic mass of Ba + 2 * atomic mass of Cl
= 137.33 g/mol + 2 * 35.45 g/mol
= 208.23 g/mol
Number of moles of BaCl2 = volume (in L) * molarity
= 0.100 L * 0.100 mol/L
= 0.0100 mol
For iron (III) sulfate (Fe2(SO4)3):
Molar mass of Fe2(SO4)3 = 2 * atomic mass of Fe + 3 * (atomic mass of S + 4 * atomic mass of O)
= 2 * 55.85 g/mol + 3 * (32.07 g/mol + 4 * 16.00 g/mol)
= 399.88 g/mol
Number of moles of Fe2(SO4)3 = volume (in L) * molarity
= 0.100 L * 0.100 mol/L
= 0.0100 mol
Since both reactants have the same number of moles (0.0100 mol), we can determine the limiting reagent by comparing the mole-to-mole ratio between BaCl2 and BaSO4 in the balanced equation.
From the equation, the mole-to-mole ratio is 3:3, or 1:1. Therefore, the limiting reagent is the one with a lesser number of moles, which is both BaCl2 and Fe2(SO4)3.
Since both reagents are the limiting reagent, we compare their molar masses to find the product with the least molar mass, which is BaSO4.
Molar mass of BaSO4 = atomic mass of Ba + atomic mass of S + 4 * atomic mass of O
= 137.33 g/mol + 32.07 g/mol + 4 * 16.00 g/mol
= 233.38 g/mol
Now, we can calculate the mass of barium sulfate formed using the limiting reagents:
Mass of BaSO4 = number of moles of limiting reagent * molar mass of BaSO4
= 0.0100 mol * 233.38 g/mol
= 2.33 g
Therefore, the mass of barium sulfate that can be produced when 100 mL of a 0.100 M solution of barium chloride is mixed with 100.0 mL of a 0.100 M solution of iron (III) sulfate is 2.33 grams.
To determine the mass of barium sulfate that can be produced, we need to calculate the limiting reagent in the reaction between barium chloride (BaCl2) and iron (III) sulfate (Fe2(SO4)3). The limiting reagent is the reactant that gets completely used up first and determines the maximum amount of product that can be formed.
The balanced chemical equation for the reaction between BaCl2 and Fe2(SO4)3 is:
3 BaCl2 + Fe2(SO4)3 → 3 BaSO4 + 2 FeCl3
From the equation, we can see that three moles of BaCl2 react with one mole of Fe2(SO4)3 to produce three moles of BaSO4. This means that the molar ratio between BaCl2 and BaSO4 is 3:3, or 1:1.
Now, let's begin by calculating the number of moles of BaCl2 and Fe2(SO4)3 in the given solutions.
For BaCl2:
Molarity (M) = moles/volume in liters
Molarity (M) = 0.100 moles/L
Volume of BaCl2 solution = 100 mL = 0.1 L
Moles of BaCl2 = Molarity * Volume
Moles of BaCl2 = 0.100 mol/L * 0.1 L
Moles of BaCl2 = 0.010 mol
For Fe2(SO4)3:
Molarity (M) = moles/volume in liters
Molarity (M) = 0.100 moles/L
Volume of Fe2(SO4)3 solution = 100.0 mL = 0.1 L
Moles of Fe2(SO4)3 = Molarity * Volume
Moles of Fe2(SO4)3 = 0.100 mol/L * 0.1 L
Moles of Fe2(SO4)3 = 0.010 mol
From the calculations, we can see that both BaCl2 and Fe2(SO4)3 have the same number of moles, suggesting that they are present in stoichiometrically equal amounts.
Therefore, the limiting reagent is BaCl2 because it will react with all of its moles, and Fe2(SO4)3 will be in excess.
Now, we can calculate the mass of BaSO4 produced from the limiting reagent:
From the balanced chemical equation, we know that 1 mole of BaCl2 produces 1 mole of BaSO4.
The molar mass of BaSO4 is:
Ba = 137.33 g/mol
S = 32.07 g/mol
O = 16.00 g/mol (four oxygen atoms in BaSO4)
Molar mass BaSO4 = 137.33 g/mol + 32.07 g/mol + (16.00 g/mol * 4)
Molar mass BaSO4 = 233.38 g/mol
Moles of BaSO4 = Moles of BaCl2 (since one-to-one molar ratio)
Moles of BaSO4 = 0.010 mol
Mass of BaSO4 = Moles of BaSO4 * Molar mass BaSO4
Mass of BaSO4 = 0.010 mol * 233.38 g/mol
Mass of BaSO4 = 2.3338 g
Therefore, approximately 2.33 grams of barium sulfate can be produced when 100 mL of 0.100 M barium chloride solution is mixed with 100.0 mL of a 0.100 M iron (III) sulfate solution.
When amounts are given for BOTH reactants you know the question is a limiting reagent (LR) problem.
3BaCl2 + Fe2(SO4)3 ==> 3BaSO4 + 2FeCl3
millimols BaCl2 = mL x M = 100 x 0.1 = 10
mmols Fe3(SO4)3 = 100 x 0.1 = 10
How much BaSO4 can be formed from the BaCl2 IF we had all of the Fe3(SO4)3 we needed.
That's 10 mmols BaCl2 x (3 mol BaSO4/3 mols BaCl2) = 10 x 1/1 = 10 mmols BaSO4
How much BaSO4 can be formed from the Fe3(SO4)2 IF we had all of the BaSO4 needed.
That's 10 mmols Fe2(SO4)3 x (3 mols BaSO4/1 mol Fe2(SO4)3) = 10 x 3/1 = 30
You see the number for BaSO4 is different which means one of them is not right; the correct value in LR problems is ALWAYS the smaller value and the reagent producing that value is the LR. So we will produce 10 mmoles (0.01 mols) BaSO4 and use all of the BaCl2. Convert 0.01 mols BaSO4 to grams. g = mols x molar mass = ?