150.0 mL of 0.105 M aluminium sulfate is mixed with 250.0 mL of 0.198 M barium chloride. What mass of barium sulfate would be precipitated?
I want to get your attention with this separate answer. I have notice over the last week or so that you are posting the SAME problem over and over. The problems are not EXACTLY the same; i.e., the numbers have been changed but the problem is the same except for the numbers. The idea in getting help on this site is for you to LEARN how to do them on you own. I fear you are not doing that and I am being used to work all/many/most of your homework.
We need to change that perception. I know I have worked at least a half dozen stoichiometry problems. Two should have been enough for you to see how to do them. All of them are alike; limiting reagent problems add a quirk to those, the molarity problems are the same. So help us help you by keeping the lines of communication open. Perhaps you should show what you have done so far in a problem and tell us what you don't understand about the problem. Good luck in you chem class/es. Chemistry isn't hard but it does require discipline.
Thank you for the update and for the information. While taking chemistry in high school helps it is not a prerequisite for doing well in chemistry at the university level. I can see that your stress level is high but you must realize that you have bitten off a large chunk to chew. One thing you could do is start a notebook for each problem that requires help, then refer to those notes when a new problem comes up. Good luck on the upcoming exam.
what is the reaction equation?
convert grams to moles, and you can tell how many moles will be produced.
convert back to grams.
Al2(SO4)3(aq) + 3BaCl2(aq) ==> 3BaSO4(s) + 2AlCl3(aq)
When two solutions are mixed like this you almost always have a limiting reagent (LR) problem.
mmols Al2(SO4)3 = 150.0 x 0.105 = 15.75
mmols BaCl2 = 250.0 x 0.198 = 49.5
How many mols can be prepared from Al2(SO4)3. That is
15.75 x (3 mols BaSO4/ 1 mol Al2(SO4)3 = 47.25
What about from 49.5 mmols BaCl2. That is
49.5 x (3 mols BaSO4/3 mols BaCl2) = 49.5
In LR problems, the SMALL number always wins; therefore you will be able to prepare 47.5 mmols BaSO4. Divide by 1000 to convert to mols, then multiply by molar mass to convert to grams.
Hey DrBob, I want to say thank you very much for helping me throughout the past 2 weeks, but the thing is I am going into university and the course I want to do requires me to do chemistry but I never did chemistry in high school so I am taking a bridging course to get me familiar with chemistry. For me, as I have never studied chemistry at this level is very difficult for me and I have 8 topics to get through in about 2 weeks. I am doing tonnes of practice questions but whenever i see a "different" one I tend to get lost in the question and get very confused. There are a lot of new concepts I have to learn and it's very annoying because I don't have a lot of time to learn this properly. I've got my exam this Friday and then I am done with chemistry but once again I am very very appreciative of your help.
No problem, but trust me I definitely do have notes on the problems I have struggled with so far and it has helped me to a medium extent but as I said whenever I see another problem that I haven't encountered before I tend to just get lost probably because of the wording. I know that it uses the concepts I have learn't but I still just struggle with it. I always do give the question a genuine attempt though.
To find the mass of barium sulfate precipitated, we first need to determine which reactant will be limiting in this reaction. The limiting reactant is the one that is completely consumed and determines the amount of product formed.
We can start by writing the balanced chemical equation for the reaction between aluminium sulfate (Al₂(SO₄)₃) and barium chloride (BaCl₂):
2Al₂(SO₄)₃ + 3BaCl₂ → 2AlCl₃ + 3BaSO₄
From the equation, we can observe that the stoichiometric ratio between aluminium sulfate and barium sulfate is 2:3. This means that for every 2 moles of aluminium sulfate, we will need 3 moles of barium sulfate.
Next, let's calculate the number of moles of each reactant:
Number of moles of aluminium sulfate (Al₂(SO₄)₃):
moles = volume (in liters) × molarity
moles = 150.0 mL × (1 L / 1000 mL) × 0.105 M
moles = 0.01575 mol
Number of moles of barium chloride (BaCl₂):
moles = volume (in liters) × molarity
moles = 250.0 mL × (1 L / 1000 mL) × 0.198 M
moles = 0.0495 mol
Now, we need to determine the limiting reactant. To do this, we compare the ratios of the moles of the reactants to their stoichiometric coefficients in the balanced equation.
For aluminium sulfate:
moles of aluminium sulfate / stoichiometric coefficient of aluminium sulfate = 0.01575 mol / 2 = 0.007875
For barium chloride:
moles of barium chloride / stoichiometric coefficient of barium chloride = 0.0495 mol / 3 = 0.0165
The lower value among these ratios represents the limiting reactant. In this case, 0.007875 is smaller than 0.0165, indicating that aluminium sulfate is the limiting reactant.
Using the stoichiometric ratio from the balanced equation, we can determine the number of moles of barium sulfate formed. Since the ratio is 2:3 between aluminium sulfate and barium sulfate, the moles of barium sulfate formed will be:
moles of barium sulfate = (moles of aluminium sulfate) × (3 / 2)
moles of barium sulfate = 0.007875 mol × (3 / 2)
moles of barium sulfate = 0.0118125 mol
Finally, we can calculate the mass of barium sulfate formed using its molar mass:
mass of barium sulfate = moles of barium sulfate × molar mass
mass of barium sulfate = 0.0118125 mol × 233.39 g/mol (molar mass of BaSO₄)
mass of barium sulfate = 2.76 g
Therefore, approximately 2.76 grams of barium sulfate would be precipitated in this reaction.