For each of the following combinations of solutions, determine the mass of barium sulfate, BaSO4, that should precipitate...?
For each of the following combinations of solutions, determine the mass of barium sulfate, BaSO4, that should precipitate. (Begin by writing a chemical equation for the reaction):
K2SO4(aq)+BaCl2(aq)=BaSO4(s)+2KCl(aq)
A) 500.0 mL of 0.100 M BaCl2 and 90.0 mL of 0.500 M K2SO4 Answer: 10.5g BaSO4
B) 100.0 mL of 0.100 M BaCl2 and 100.0 mL of 0.500 M K2SO4 Answer: 2.33g BaSO4
C) 100.0 mL of 0.100 M BaCl2 and 500.0 mL of 0.500 M K2SO4 Answer: 2.33g BaSO4
I need to know how to solve this problem, but I don't understand how to actually do it.
For A, I started with finding the mol of BaCl2, which is 0.0500 mol BaCl2, and for K2SO4, 0.045 mol K2SO4.
Then I use Stoichiometry calculations:
0.05 mol BaCl2*1 mol BaSO4/1 mol BaCl2* 208.2g BaSO4/1 mol BaSO4= 10.4g BaSO4.
Then I did the other information given, because I was given information about K2SO4 (mainly because I don't know what I am doing and if I should actually do it.):
0.045 mol K2SO4*1 mol BaSO4/1 mol K2SO4*208.2g BaSO4/1 mol BaSO4= 9.37g BaSO4
For B, 0.01 mol BaCl2 and 0.05 mol K2SO4.
0.01 mol BaCl2*1 mol BaSO4/1 mol BaCl2*208.2g BaSO4/1 mol BaSO4= 2.08g BaSO4
0.05 mol K2SO4*1 mol BaSO4/ 1 mol K2SO4*208.2g BaSO4/1 mol BaSO4= 10.4g BaSO4
I am doing something wrong, maybe I missed a step or something. Do I really need to use both of the information given? Because this is really confusing the hell out of me. Textbook: ISBN:978-0-07-340267-3
A is a limiting reagent (LR) problem. You know that when you are given amounts for BOTH reactants. You use BOTH to determine which is the limiting reagent, then use the one that is the LR.
Right. You have 0.05 mol BaCl2 and you have 0.045 mols K2SO4.
a. Convert mols BaCl2 to mols BaSO4. Since that's 1:1 you could get 0.05 mols BaSO4 IF you had all of the K2SO4 you need to do that.
b. Convert mols K2SO4 to mols BaSO4. You can get 0.045 mols BaSO4 from that IF you had all of the BaCl2 needed to do that.
c. in LR problems, the correct value of BaSO4 you get is ALWAYS the smaller number because, in this case, the K2SO4 is the limiting reagent. So you know you get 0.045 mols BaSO4.
d. Convert to grams. mols BaSO4 x molar mass BaSO4 = grams BaSO4; i.e., 0.045 x 233.4 = 10.5 grams.
I didn't work through the others but probably they are the same type.
To solve the problem and find the mass of barium sulfate (BaSO4) that should precipitate, you need to use the stoichiometry of the balanced equation and the given concentrations and volumes of the solutions. Here is a step-by-step solution for each combination of solutions:
A) 500.0 mL of 0.100 M BaCl2 and 90.0 mL of 0.500 M K2SO4:
1. Start by finding the number of moles of BaCl2 in 500.0 mL:
moles of BaCl2 = volume (L) × concentration (M)
moles of BaCl2 = (500.0 mL ÷ 1000 mL/L) × 0.100 M
moles of BaCl2 = 0.0500 mol BaCl2
2. Find the number of moles of K2SO4 in 90.0 mL:
moles of K2SO4 = volume (L) × concentration (M)
moles of K2SO4 = (90.0 mL ÷ 1000 mL/L) × 0.500 M
moles of K2SO4 = 0.0450 mol K2SO4
3. Use the balanced equation to relate the moles of BaCl2 to the moles of BaSO4:
From the balanced equation: 1 mol BaCl2 reacts with 1 mol BaSO4
moles of BaSO4 = moles of BaCl2
4. Calculate the mass of BaSO4 precipitated using the molar mass of BaSO4:
mass of BaSO4 = moles of BaSO4 × molar mass of BaSO4
mass of BaSO4 = 0.0500 mol × 208.2 g/mol
mass of BaSO4 = 10.4 g BaSO4
B) 100.0 mL of 0.100 M BaCl2 and 100.0 mL of 0.500 M K2SO4:
1. Find the number of moles of BaCl2 in 100.0 mL:
moles of BaCl2 = (100.0 mL ÷ 1000 mL/L) × 0.100 M
moles of BaCl2 = 0.0100 mol BaCl2
2. Find the number of moles of K2SO4 in 100.0 mL:
moles of K2SO4 = (100.0 mL ÷ 1000 mL/L) × 0.500 M
moles of K2SO4 = 0.0500 mol K2SO4
3. Use the balanced equation to relate the moles of BaCl2 to the moles of BaSO4:
From the balanced equation: 1 mol BaCl2 reacts with 1 mol BaSO4
moles of BaSO4 = moles of BaCl2
4. Calculate the mass of BaSO4 precipitated using the molar mass of BaSO4:
mass of BaSO4 = moles of BaSO4 × molar mass of BaSO4
mass of BaSO4 = 0.0100 mol × 208.2 g/mol
mass of BaSO4 = 2.08 g BaSO4
C) 100.0 mL of 0.100 M BaCl2 and 500.0 mL of 0.500 M K2SO4:
1. Find the number of moles of BaCl2 in 100.0 mL:
moles of BaCl2 = (100.0 mL ÷ 1000 mL/L) × 0.100 M
moles of BaCl2 = 0.0100 mol BaCl2
2. Find the number of moles of K2SO4 in 500.0 mL:
moles of K2SO4 = (500.0 mL ÷ 1000 mL/L) × 0.500 M
moles of K2SO4 = 0.250 mol K2SO4
3. Use the balanced equation to relate the moles of BaCl2 to the moles of BaSO4:
From the balanced equation: 1 mol BaCl2 reacts with 1 mol BaSO4
moles of BaSO4 = moles of BaCl2
4. Calculate the mass of BaSO4 precipitated using the molar mass of BaSO4:
mass of BaSO4 = moles of BaSO4 × molar mass of BaSO4
mass of BaSO4 = 0.0100 mol × 208.2 g/mol
mass of BaSO4 = 2.08 g BaSO4
Note: In each case, there is only one limiting reactant (BaCl2) and the moles of BaSO4 formed are equal to the moles of BaCl2. The answer you provided was correct for case A, but there was a slight error in case B. The correct mass of BaSO4 for case B is 2.08 g, not 10.4 g.
To solve this problem, you are on the right track with your calculations. Let's break it down step by step:
First, write the balanced chemical equation:
K2SO4(aq) + BaCl2(aq) = BaSO4(s) + 2KCl(aq)
This equation shows that 1 mole of BaCl2 reacts with 1 mole of BaSO4.
For part A, you correctly found the moles of BaCl2 and K2SO4. However, there is an error in your calculation. Let's rework it correctly:
0.0500 mol BaCl2 * (1 mol BaSO4/1 mol BaCl2) * (208.2 g BaSO4/1 mol BaSO4) = 10.4 g BaSO4
So the correct mass of BaSO4 that should precipitate for Part A is 10.4 g, not 10.5 g.
Now, for the information about K2SO4, you should ignore it because the question asks for the mass of BaSO4 precipitated, not K2SO4.
In Part B, your calculations were correct, but you made a mistake in your interpretation of the result. You found the mass of BaSO4 for BaCl2 as 2.08 g and for K2SO4 as 10.4 g. However, remember that you are determining the mass of BaSO4 that should precipitate, meaning the actual mass of BaSO4 formed from the limiting reactant. In this case, the limiting reactant is BaCl2 because its amount is lesser.
So the correct mass of BaSO4 that should precipitate for Part B is 2.08 g, not 2.33 g.
Again, for the information about K2SO4, you can ignore it.
For Part C, you need to repeat the same calculations as Part B, considering BaCl2 as the limiting reactant.
So the correct mass of BaSO4 that should precipitate for Part C is also 2.08 g, not 2.33 g.
In summary:
- For Part A, the mass of BaSO4 that should precipitate is 10.4 g.
- For Parts B and C, the mass of BaSO4 that should precipitate is 2.08 g.
Remember to always consider the limiting reactant when calculating the mass of the precipitate.