500ml of 0.250M Na2SO4 solution is added to an aqueous solution of 15.00 g of BaCl2 resulting in the formation of white precipitate of BaSO4. How many moles and how many grams of BaSO4 are formed?
This is a limiting reagent problem.
Na2SO4 + BaCl2 ==> BaSO4 + 2NaCl
mols Na2SO4 = M x L = ?
mols BaCl2 = grams/molar mass = ?
Convert mols Na2SO4 to mols BaSO4 using the coefficients in the balanced equation.
Convert mols BaCl2 to mols BaSO4--same process.
It is likely you will get two different values for mols BaSO4. The correct value in limiting reagent problems is ALWAYS the smaller one and the reagent providing that value is the limiting reagent.
Then grams BaSO4 = mols BaSO4 (the smaller value) x molar mass BaSO4.
To find the number of moles of BaSO4 formed, we need to first calculate the number of moles of BaCl2, and then use the stoichiometry of the balanced chemical equation to determine the number of moles of BaSO4.
Step 1: Calculate the number of moles of BaCl2.
Given:
Mass of BaCl2 = 15.00 g
Molar mass of BaCl2 = 137.33 g/mol
Number of moles of BaCl2 = Mass of BaCl2 / Molar mass of BaCl2
= 15.00 g / 137.33 g/mol
≈ 0.1093 mol
Step 2: Use the stoichiometry of the balanced chemical equation to determine the number of moles of BaSO4 formed.
The balanced chemical equation for the reaction is:
Na2SO4 + BaCl2 -> BaSO4 + 2NaCl
According to the balanced equation, 1 mole of BaCl2 produces 1 mole of BaSO4. Therefore, the number of moles of BaSO4 formed will be the same as the number of moles of BaCl2.
Number of moles of BaSO4 formed ≈ 0.1093 mol
Step 3: Calculate the mass of BaSO4 formed.
Given:
Molar mass of BaSO4 = 233.38 g/mol
Mass of BaSO4 = Number of moles of BaSO4 formed * Molar mass of BaSO4
= 0.1093 mol * 233.38 g/mol
≈ 25.46 g
Therefore, approximately 0.1093 moles and 25.46 grams of BaSO4 are formed.
To find the number of moles of BaSO4 formed, we need to use stoichiometry and the balanced chemical equation for the reaction between BaCl2 and Na2SO4:
BaCl2 + Na2SO4 -> BaSO4 + 2NaCl
From the balanced equation, we can see that 1 mole of BaCl2 reacts with 1 mole of Na2SO4 to form 1 mole of BaSO4.
First, let's calculate the number of moles of BaCl2:
The molar mass of BaCl2 = 137.33 g/mol (from periodic table)
The mass of BaCl2 = 15.00 g (given)
Number of moles = mass / molar mass
Number of moles of BaCl2 = 15.00 g / 137.33 g/mol
Number of moles of BaCl2 ≈ 0.1093 mol
Since the molar ratio between BaSO4 and BaCl2 is 1:1, the number of moles of BaSO4 formed is also approximately 0.1093 mol.
To find the mass of BaSO4 formed, we can use the molar mass of BaSO4:
The molar mass of BaSO4 = 233.38 g/mol (from periodic table)
Mass of BaSO4 = number of moles × molar mass
Mass of BaSO4 = 0.1093 mol × 233.38 g/mol
Mass of BaSO4 ≈ 25.45 g
Therefore, approximately 0.1093 moles and 25.45 grams of BaSO4 are formed.