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
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.
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.
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 = ?
I can help you with your problems! Let's work through each one step by step.
Problem 1: Finding the grams of barium sulfate formed.
To solve this problem, we need to use the balanced chemical equation provided, which is:
BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaCl(aq)
First, let's calculate the number of moles of barium chloride (BaCl2) and sodium sulfate (Na2SO4) used in the reaction. Use the given volumes and molarities:
Moles of BaCl2 = volume (L) of BaCl2 solution × molarity of BaCl2
Moles of Na2SO4 = volume (L) of Na2SO4 solution × molarity of Na2SO4
Next, we need to determine which reactant is the limiting one. This can be done by comparing the molar ratios between BaCl2 and Na2SO4 in the balanced equation. The reactant with a smaller molar ratio will be the limiting reactant.
Once the limiting reactant is determined, we can use the molar ratio between BaCl2 and BaSO4 in the balanced equation to calculate the moles of BaSO4 formed.
Finally, we convert the moles of BaSO4 to grams using the molar mass of BaSO4.
Try following these steps and let me know if you're still having trouble.
Problem 2: Concentration of KBr in a solution prepared by mixing.
To find the concentration of KBr in the final solution, we can use the formula:
Cfinal = (C1 × V1 + C2 × V2) / (V1 + V2)
where C1 and C2 are the initial concentrations of KBr solutions, and V1 and V2 are the initial volumes.
Substitute the given values into the formula and calculate the final concentration, Cfinal.
Remember to convert all volumes to liters if they are given in milliliters.
If you've tried this approach and are still experiencing difficulties, please let me know.
Problem 3: Molarity of sodium ion in the solution.
To find the molarity of sodium ions (Na+) in the solution, we first need to calculate the initial moles of sodium chloride and sodium sulfate using the given volume and molarity.
Moles of NaCl = volume (L) of NaCl solution × molarity of NaCl
Moles of Na2SO4 = volume (L) of Na2SO4 solution × molarity of Na2SO4
Next, we determine the moles of sodium ions contributed by NaCl and Na2SO4 separately. This is done by multiplying the moles of each compound by the number of sodium ions in the respective chemical formulas.
Finally, we add the moles of sodium ions together and divide by the total volume of the solution in liters (converted from milliliters). This gives us the molarity of sodium ions in the solution.
I hope this helps! If you have any further questions or need more specific guidance, please let me know.