How much of a 1.50 M sodium sulfate solution in milliliters is required to completely precipitate all of the barium in 150.0 mL of a 0.250 M barium nitrate solution?
Problem 118P :
Molarity of Sodium sulfate(Na2SO4) solution = 1.50 M
Molarity of barium nitrate(Ba(NO3)2) solution = 0.250 M
Volume of silver nitrate(AgNO3) solution = 150.0 mL
Volume of solution is in mL, so let’s convert to L :
1 L = 1000 mL
Therefore, 150.0 mL in L will be :
= 150 mL
= 0.150 L.
First, Let’s write the chemical reaction that’s happening :
Na2SO4 (aq) + Ba(NO3)2 (aq) → 2NaNO3 (aq) + BaSO4 (s)
Now, let’s find the number of moles of barium nitrate( Ba(NO3)2) using the molarity formula as barium nitrate is the solute :
Therefore, the number of moles(m) can be derived from the above formula as :
Moles of solute = Molarity Volume of Solution (in L)
n = 0.250 M0.150 L
n = 0.0375 moles
Textbook: Introductory Chemistry
Author: Nivaldo J Tro
Since the solution to 118P from 13 chapter was answered, more than 557 students have viewed the full step-by-step answer. This full solution covers the following key subjects: barium, solution, precipitate, much, nitrate. This expansive textbook survival guide covers 19 chapters, and 2046 solutions. The answer to “How much of a 1.50 M sodium sulfate solution in milliliters is required to completely precipitate all of the barium in 150.0 mL of a 0.250 M barium nitrate solution?” is broken down into a number of easy to follow steps, and 30 words. Introductory Chemistry was written by and is associated to the ISBN: 9780321910295. The full step-by-step solution to problem: 118P from chapter: 13 was answered by , our top Chemistry solution expert on 05/06/17, 06:45PM. This textbook survival guide was created for the textbook: Introductory Chemistry, edition: 5.