Question

Please help! I have spent hours trying to work these three problems.

How many grams of solid barium sulfate form when 27.1 mL of 0.160 M barium chloride reacts with 53.4 mL of 0.065 M sodium sulfate? Aqueous sodium chloride forms also.

I have converted the 27.1mL to .0271L, and the 53.4mL to .0534L. I tried taking the 0.160M / .0271L and 0.065M / .0534L and then dividing and multiplying by taking the molar mass of barium sulfate but I cant get the answer to come out


Assuming that the volumes are additive, what is the concentration of KBr in a solution prepared by mixing 0.226 L of 0.061 M KBr with 0.584 L of 0.054 M KBr?

I've tried multiplying and dividing the Molarity by the liters but I cant seem to get it to come out. I think I am missing a step. I have tried it so many different ways I just need someone to walk me though the steps of how to work it.




Molarity of sodium ion in a solution made by mixing 3.47 mL of 0.275 M sodium chloride with 500. mL of 6.51 10-3 M sodium sulfate (assume volumes are additive)
I converted the 3.47mL to 0.0034L and the 500mL to .5000L I just don't know how to work it from there I have tried dividing multiplying different ways.



Thank you

Answer 1

This is a limiting reagent problem; you know that because amounts are given for both reactants.

How many grams of solid barium sulfate form when 27.1 mL of 0.160 M barium chloride reacts with 53.4 mL of 0.065 M sodium sulfate? Aqueous sodium chloride forms also.
BaCl2 + Na2SO4 ==> BaSO4 + 2NaCl

When I say about it means I've estimated. You should redo the problem to obtain more exact values.
mols BaCl2 = M x L = about 0.00434
mols Na2SO4 = M x L = about 0.00347

Now convert mols of each to mols BaSO4.
First BaCl2:
0.00434 x (1 mol BaSO4/1 mol BaCl2) = 0.00434 mols BaSO4 if we had 0.00434 mols BaCl2 and all of the Na2SO4 we needed.
Next Na2SO4:
0.00347 x (1 mol BaSO4/2 mols Na2SO4) = about 0.0017 mols BaSO4.

You see you have two values for mols BaSO4 which means one of them must be wrong; the correct value in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
So g BaSO4 = about 0.0017 x molar mass BaSO4.

Answer 2

Assuming that the volumes are additive, what is the concentration of KBr in a solution prepared by mixing 0.226 L of 0.061 M KBr with 0.584 L of 0.054 M KBr?

Same rules. I've estimated. You need to redo the whole thing.

mols KBr(soln 1) = M x L = 0.014
mols KBr(soln 2) = 0.032

Total mols = 0.032+0.014 = ? total mols
Total volume(if additive) = 0.226L + 0.054L = ? total L
(KBr) = total mols/total L.

Answer 3

Molarity of sodium ion in a solution made by mixing 3.47 mL of 0.275 M sodium chloride with 500. mL of 6.51 10-3 M sodium sulfate (assume volumes are additive)

Same rules about estimation. You need to redo the whole thing.

mols Na^+ in NaCl = M x L = about 0.00095
mols Na^+ in Na2SO4 = 2*M x L = about 0.00651.
total mols Na^+ = you do it.
total L = you do it.
(Na^+) = total mols Na^+/total L = ?