# Write a balanced chemical equation for the reaction and calculate the mass of BaSO4 precipitated?

Hi Everyone!

I am having the hardest time working through this problem. I used to know it at one point, and I cannnot remember what to do for the life of me. If someone could help me work through it, I would greatly greatly appreciate it. Thank you so much!

-Andrea

Exactly 750.0 mL of a solution that contained 480.4 ppm Ba(NO3)2 was mixed with 200.0mL of a 0.03090 M solution of Al2(SO4)3. BaSO4 precipitates and settles out of solution.

(a) Write a balanced Chemical equation, with phase symbols for the overall reaction taking place here.

(b) Assume the reaction goes all the way to the right. What mass (in grams) of BaSO4 is theoretically possible?

(c) Find the concentrations of each of the ions left in solution after precipitation has finished: [Ba2+], [SO4 2-], [Al3+], [NO3-]. Assume all volumes are additive.

kyle1990

Tue, 2008-09-23 10:35

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## Re: Immediate homework help!

first write the balanced equation. This is critical in solving the rest of the problem

3Ba(NO

_{3})_{2}_{(aq)}+ Al_{2}(SO_{4})_{3}_{(aq)}-----> 3BaSO_{4}_{(s)}+ 2Al(NO_{3})_{3}_{(aq)}---------------------------------------------------------------------------

now for part B

What is the maximum theoretical yield of BaSO4?

Solution:

---First convert all your concentrations into molarity (mol/L). The concentration of aluminum sulfate is already in this unit so we will leave it alone for the time being.

So we have to now convert 480.4ppm Ba(NO3)2 into moles of Ba(NO3)2. But how?

well, by definition,

1 gram of solute = 1ppm

1.0X10

^{6}g of solutionso therefore, 480.4 ppm of Ba(NO3)2= 480.4g per 1,000,000g of solution. We will assume that a million grams of solution is equal to a million mL of solution (1000L), since the density of water is about 1.0g/cm

^{3}.We must now convert 480.4 g of barium nitrate into moles:

480.4 g = 1.835mol of Ba(NO3)2

261.74g/mol

Now we must obtain the molar concentration of Ba(NO3)2:

Molarity= mol solute/L of solution

=1.835 mol/1000.L = 0.001835M

--------------------------------------------------------------------------

Both solutions are now in mol/L. Now we can calculate how many moles of Ba(NO3)2 are in 750.0 ml of solution:

Molarity X volume = moles

(.001835M)(.7500L)=0.001376 moles of Ba(NO3)2

And also we need to find out how many moles of Al2(SO4)3 are in 200.0mL of solution

Molarity X volume =moles

(.03090M)(.2000L)= .006180 moles Al2(SO4)3

-------------------------------------------------------------------------

These values we just obtained are the number of moles that we will mix together. But first we must figure out which substance is the limiting reagent-which one runs out first. Using the equation we use the mole ratio of Al2(SO4)3 to Ba(NO3)2:

1 mol Al2(SO4)3 = .006180 mol Al2(SO4)3

3 mol Ba(NO3)2 x mol of Ba(NO3)2

solving for x

x=.01854 mol of Ba(NO3)2

So this means that .01854 moles of Ba(NO3)2 are required to completely react with .006180 moles of Al2(SO4)3. But we only have .001376 moles of Ba(NO3)2. Therefore, Ba(NO3)2 is the limiting reagent. What does this mean? It means that ALL of the Ba(NO3)2 will be consumed, and only some of the Al2(SO4)3 will be used up.

And because of this, we have to do the following mole ratio to find out how many moles of BaSO4 can be formed from .001376 moles of Ba(NO3)2. But if you look at the equation, the number of moles of Ba(NO3)2 are the same as the number of moles in BaSO4. So as .001376 mol of Ba(NO3)2 are consumed, .001376 mol of BaSO4 are produced. So just convert .001376 mol of BaSO4 into grams of BaSO4:

(.001376 mol BaSO4)(233.34g/mol)= .3211 grams of BaSO4 are theoretically produced.

I hope this gave you a massive leap of help. I'll leave the last part for you to figure out. If you get stuck or don't understand how to even start it, just let me know.