Question

How can I solve for this ? 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.

Answer 1

Hi,
u have to get the stoichiometry of this reaction through a balance equation....
then u have to use that ratio with the given data and u could find the mass of precipitate.....

Answer 2

first write a reaction, determine what the precipitate is.
Find the concentration of the species which is insoluble (i.e. precipitate).

Write a Ksp equation, and find the molar solubility. Find how much you have left after you've exhausted this capacity. Convert the moles to grams.

Answer 3

this is the reaction which is happening here...
Ba(OH)2 + Na2(SO4) = Ba(SO4) + 2 Na(OH)
here one mole of Ba(OH)2 will react with one mole of Na2(SO)4...
which mean the substance which have lesser amount of moles will be the limiting agent..

no.of moles of Na2(SO4) = 3060 x 0.0664 x 10^(-3) =203.184 x 10^(-3)
no. of moles of Ba(OH)2 = 2270 x 0.082 x 10^(-3) = 186.14 x 10^(-3)

here the amount of moles of Na2(SO4) is higher than the amount of mole of Ba(OH)2...So
excess amount of Na2(SO4) will remain even after the reaction....

Answer 4

but the question asks the mass of precipitation forms...... which mean they r asking the mass of Ba(SO4) that will form in this reaction...

From the balance equation...
if there was 1 mole of Ba(OH)2 ----> 1 mole of Ba(SO4) will form

From the data given in the question...

there 186.14 x 10^(-3) moles of ----> There will be 186.14 x 10^(-3) moles of Ba(SO4)
Ba(OH)2

so the no. of moles of Ba(SO4) that will precipitate is 186.14 x 10^(-3)

to find the mass u have to multiply it by the relative atomic mass of Ba(SO4)
=233 x 186.14 x 10^(-3) g
= 43370.62 x 10^(-3) g
= 43.37062 g

hope this will help ya!!

Answer 5

Thanks so much

Answer 6

u r welcome!!

Answer 7

good stuff. Although, you're not taking into account the Ksp of BaSO4, but it's still a good approximation because it's pretty insoluble.