Calculate the mass of the precipitate formed when 2.27 L of 0.0820 M Ba(OH)2 are mixed with 3.06 L of 0.0664 M Na2SO4.
This is a limiting reagent (LR) problem. One reagent is the LR, the other is the excess reagent (ER). I do these the long way; i.e., there are shorter ways to do them but I have trouble explaining them.
Ba(OH)2 + Na2SO4 ==> BaSO4 + 2NaOH
1. mols Ba(OH)2 = M x L = ?
2. mols Na2SO4 ==> M x L = ?
3. Using the coefficients in the balanced equation, convert mols Ba(OH)2 to mols BaSO4.
4. Do the same to convert mols Na2SO4 to mols BaSO4.
5. You can see that mols BaSO4 formed is not the same for both 1 and 2. In LR problems, the SMALLER number is ALWAYS the correct for the product formed the reagent producing that number is the LR.
6. Convert the smaller number into grams. g BaSO4 = mols BaSO4 x molar mass BaSO4.
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It appears that you may need more help or clarification. Is there a particular part of the problem or solution that you are unsure about?
To calculate the mass of the precipitate formed when 2.27 L of 0.0820 M Ba(OH)2 are mixed with 3.06 L of 0.0664 M Na2SO4, we need to determine the limiting reactant in the reaction. The limiting reactant is the substance that gets completely consumed in a chemical reaction, thereby limiting the amount of product formed.
Step 1: Write the balanced equation for the reaction.
Ba(OH)2 + Na2SO4 -> BaSO4 + 2NaOH
Step 2: Convert the given volumes of the solutions to moles.
Molarity (M) = moles of solute / liters of solution
moles of solute = Molarity × liters of solution
For Ba(OH)2:
moles of Ba(OH)2 = 0.0820 M × 2.27 L = 0.18614 moles
For Na2SO4:
moles of Na2SO4 = 0.0664 M × 3.06 L = 0.20304 moles
Step 3: Determine the stoichiometry of the reaction to find the ratio of moles between Ba(OH)2 and Na2SO4. From the balanced equation, we can see that the ratio is 1:1.
Step 4: Compare the moles of Ba(OH)2 and Na2SO4 to identify the limiting reactant.
In this case, both reactants have a 1:1 ratio, so the reactant with fewer moles is the limiting reactant. In this case, Ba(OH)2 has 0.18614 moles and Na2SO4 has 0.20304 moles. Thus, Ba(OH)2 is the limiting reactant.
Step 5: Calculate the moles of the precipitate formed using the stoichiometry of the balanced equation.
Since Ba(OH)2 is the limiting reactant, the number of moles of BaSO4 formed is equal to the number of moles of Ba(OH)2 used.
moles of BaSO4 formed = 0.18614 moles
Step 6: Calculate the molar mass of BaSO4 to convert moles to grams.
The molar mass of BaSO4 is:
(1 × atomic mass of Ba) + (1 × atomic mass of S) + (4 × atomic mass of O)
= (1 × 137.33 g/mol) + (1 × 32.07 g/mol) + (4 × 16.00 g/mol)
= 233.38 g/mol
Step 7: Calculate the mass of the precipitate formed.
mass of BaSO4 formed = moles of BaSO4 formed × molar mass of BaSO4
mass of BaSO4 formed = 0.18614 moles × 233.38 g/mol
Therefore, the mass of the precipitate formed when 2.27 L of 0.0820 M Ba(OH)2 are mixed with 3.06 L of 0.0664 M Na2SO4 is approximately 43.39 grams.