Hi, I need help figuring out this question.
An impure sample of barium hydroxide, Ba(OH)2 (aq), has a mass of 0.540 g. It is dissolved in water and then treated with excess sulfuric acid, H2SO4(aq). This results in the formation of a precipitate of barium sulfate, BaSO4(s).
Given balanced chemical equation:
H2SO4(aq) + Ba(OH)2(aq) --> BaSO4(s) + H2O(l)
The barium sulfate is filtered, and any remaining sulfuric acid is washed away. Then the barium sulfate is dried and its mass is measured to be
0.62 g. What mass of barium hydroxide was in the original (impure) sample?
//I know 0.62 g is supposed to be the actual yield. I'm guessing that I am supposed to find the theoretical yield of barium sulfate and use that to find barium hydroxide but I can't figure out how to get there.
To find the mass of barium hydroxide in the original sample, we need to use stoichiometry.
Step 1: Determine the molar mass of barium hydroxide
Ba(OH)2 = (137.33 g/mol) + 2(16.00 g/mol) + 2(1.01 g/mol) = 171.34 g/mol
Step 2: Convert the given mass of barium sulfate to moles
0.62 g BaSO4 x (1 mol BaSO4 / 233.38 g BaSO4) = 0.0026561 mol BaSO4
Step 3: Use the stoichiometry of the balanced chemical equation to determine the moles of Ba(OH)2
From the balanced equation, we see that 1 mol BaSO4 is produced from 1 mol Ba(OH)2.
So, the moles of Ba(OH)2 = 0.0026561 mol BaSO4
Step 4: Convert the moles of Ba(OH)2 to mass
Mass of Ba(OH)2 = (0.0026561 mol Ba(OH)2) x (171.34 g/mol Ba(OH)2) = 0.4545 g Ba(OH)2
Therefore, the mass of barium hydroxide in the original impure sample is approximately 0.4545 g.
To find the mass of barium hydroxide in the original sample, we need to determine the amount of barium sulfate formed and then relate it to the amount of barium hydroxide used in the reaction. Here's the step-by-step approach to solve this problem:
1. Calculate the molar mass of BaSO4:
The molar mass of barium sulfate, BaSO4, can be determined by adding the atomic masses of its constituent elements: Ba (barium) = 137.33 g/mol, S (sulfur) = 32.06 g/mol, and O (oxygen) = 16.00 g/mol. Adding these gives: 137.33 + 32.06 + (16.00 × 4) = 233.39 g/mol.
2. Convert the mass of the barium sulfate obtained (0.62 g) into moles:
Divide the mass of barium sulfate by its molar mass: 0.62 g ÷ 233.39 g/mol ≈ 0.00266 mol.
3. Use the balanced chemical equation to find the molar ratio between barium sulfate and barium hydroxide:
The stoichiometric coefficients in the balanced chemical equation indicate the mole ratio between reactants and products. From the equation: 1 mol of BaSO4 is produced from 1 mol of Ba(OH)2.
4. Determine the moles of barium hydroxide in the original sample:
Since the mole ratio is 1:1, the moles of barium hydroxide are also approximately 0.00266 mol.
5. Calculate the mass of barium hydroxide:
To find the mass of barium hydroxide, multiply the number of moles by its molar mass. The molar mass of Ba(OH)2 can be calculated as: Ba (barium) = 137.33 g/mol, O (oxygen) = 16.00 g/mol, and H (hydrogen) = 1.01 g/mol. Adding these gives: 137.33 + (16.00 × 2) + (1.01 × 2) = 171.35 g/mol.
Therefore, the mass of barium hydroxide in the original sample is approximately: 0.00266 mol × 171.35 g/mol ≈ 0.456 g.
Hence, the mass of barium hydroxide in the original impure sample is approximately 0.456 g.
find the moles of barium in the sulfate
... these are the moles that were in the original (impure) sample
use the moles to find the moles (and mass) of the barium hydroxide