What mass of barium sulphate can be produced when 100.0 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) sulphate?
Please explain!
moles BaCl2 = M x L = 0.100 x 0.100L = ?
moles Fe2(SO4)3 = M x L = ?
..3BaCl2 + Fe2(SO4)3 ==> 3BaSO4 + 2FeCl3
You have the initial amount from above.
Convert mole BaCl2 to moles BaSO4
Convert moles Fe2(SO4)3 to mols BaSO4.
The SMALLER number is ALWAYS the number to choose when you have a limiting reagent problem (and we know it is a limiting reagent because amounts for BOTH reactants are given).
Then grams = moles BaSO4 x molar mass BaSO4.
Okay this makes sense thank you!
To find the mass of barium sulphate, we need to use stoichiometry and the balanced chemical equation for the reaction between barium chloride (BaCl2) and iron (III) sulphate (Fe2(SO4)3) to form barium sulphate (BaSO4).
The balanced chemical equation is:
3 BaCl2 + Fe2(SO4)3 → 3 BaSO4 + 2 FeCl3
From the equation, we can see that 3 moles of barium chloride react with 1 mole of iron (III) sulphate to produce 3 moles of barium sulphate. This means that the ratio of barium chloride to barium sulphate is 3:3 (or 1:1).
First, let's calculate the number of moles of barium chloride (BaCl2) and iron (III) sulphate (Fe2(SO4)3) in the solutions:
Number of moles of BaCl2 = volume (in liters) × molarity
Number of moles of BaCl2 = 0.1 L × 0.1 M = 0.01 moles
Number of moles of Fe2(SO4)3 = volume (in liters) × molarity
Number of moles of Fe2(SO4)3 = 0.1 L × 0.1 M = 0.01 moles
Since the stoichiometry ratio between BaCl2 and BaSO4 is 1:1, we can say that 0.01 moles of BaCl2 will produce 0.01 moles of BaSO4.
Next, we need to calculate the molar mass of BaSO4:
Ba = 137.33 g/mol
S = 32.06 g/mol
O = 16.00 g/mol (4 oxygen atoms in BaSO4)
Molar mass of BaSO4 = (137.33 g/mol) + (32.06 g/mol) + (4 × 16.00 g/mol) = 233.38 g/mol
Now, let's calculate the mass of barium sulphate (BaSO4) using the mole-to-mass conversion:
Mass of BaSO4 = number of moles × molar mass
Mass of BaSO4 = 0.01 moles × 233.38 g/mol = 2.33 g
Therefore, the mass of barium sulphate that can be produced when 100.0 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) sulphate is 2.33 grams.
To determine the mass of barium sulphate that can be produced when the two solutions are mixed, we need to use the concept of stoichiometry and balanced chemical equations.
Step 1: Write the balanced chemical equation.
The balanced chemical equation for the reaction between barium chloride (BaCl2) and iron (III) sulphate (Fe2(SO4)3) can be written as follows:
3BaCl2 + Fe2(SO4)3 → 3BaSO4 + 2FeCl3
Step 2: Determine the limiting reagent.
To find the limiting reagent, we compare the number of moles of each reactant used in the reaction. The limiting reagent is the one that is completely consumed and determines the amount of product formed.
Let's calculate the number of moles for each reactant:
Number of moles of BaCl2 = volume (in liters) × molarity = (100.0 mL / 1000 mL/L) × 0.100 M = 0.0100 mol
Number of moles of Fe2(SO4)3 = volume (in liters) × molarity = (100.0 mL / 1000 mL/L) × 0.100 M = 0.0100 mol
Based on the balanced chemical equation, we can determine the stoichiometric ratio between BaCl2 and BaSO4, as well as between Fe2(SO4)3 and BaSO4. The ratio is 3:3, which means that one mole of BaCl2 reacts with one mole of Fe2(SO4)3 to produce one mole of BaSO4.
Since the number of moles of both reactants is the same, neither is in excess, and therefore, neither is the limiting reagent. This means that the balanced chemical equation allows for the complete reaction of both reactants.
Step 3: Calculate the mass of BaSO4 produced.
The molar mass of BaSO4 can be calculated as:
Molar mass of BaSO4 = atomic mass of Ba + atomic mass of S + 4 × atomic mass of O
= (1 × 137.33 g/mol) + (1 × 32.07 g/mol) + (4 × 16.00 g/mol)
= 137.33 g/mol + 32.07 g/mol + 64.00 g/mol
= 233.40 g/mol
The theoretical yield of BaSO4 can be calculated by multiplying the number of moles of BaSO4 by its molar mass:
Theoretical yield of BaSO4 = number of moles of BaSO4 × molar mass of BaSO4
= 0.0100 mol × 233.40 g/mol
= 2.334 g
Therefore, when 100.0 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) sulphate, the mass of barium sulphate produced will be 2.334 grams.