Given the balanced equation: HBr(aq) + NaOH(aq) NaBr(aq) + H2O) and remembering that Molarity = moles/liter OR mmoles/mL (1) (a) Calculate the number of mmoles of HBr in 100.0 mL of 0.250 MHBR (b) Calculate the number of mmoles of NaOH in 100.0 mL of 0.250 MNAOH © When these two solutions are mixed the acid and base should neutralize one another exactly. This means that all of the acid and base are completely used up; either one could be considered a "limiting reactant". Starting with the mmoles of either the acid or base, calculate the number of mmoles of salt produced by the reaction.
Given the balanced equation: HBr(aq) + NaOH(aq) NaBr(aq) + H2O) and remembering that Molarity = moles/liter OR mmoles/mL (1) (a) Calculate the number of mmoles of HBr in 100.0 mL of 0.250 MHBR (b) Calculate the number of mmoles of NaOH in 100.0 mL of 0.250 MNAOH © When these two solutions are mixed the acid and base should neutralize one another exactly. This means that all of the acid and base are completely used up; either one could be considered a "limiting reactant". Starting with the mmoles of either the acid or base, calculate the number of mmoles of salt produced by the reaction.
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![Given the balanced equation:
HBr(aq) + NaOH(aq)
NaBr(ag) + H2O()
and remembering that
Molarity = moles/liter
OR
mmoles/mL
(1) (a) Calculate the number of mmoles of HBr in 100.0 mL of 0.250 MHB.
(b) Calculate the number of mmoles of NaOH in 100.0 mL of 0.250 MNAOH
©) When these two solutions are mixed the acid and base should neutralize one
another exactly. This means that all of the acid and base are completely used up;
ing reactant". Starting with the mmoles of
either the acid or base, calculate the number of mmoles of salt produced by the
either one could be considered a "li
reaction.
(d) Using the mmoles of salt produced and the total volume in mL of solution (from the
mixing of the acid and base solutions), calculate the molarity of the salt solution
produced by this reaction.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa389d800-2220-480d-b079-1654ac8487de%2Fc7044113-ed79-4672-854f-f21d9eba53f2%2Fqxpz45it_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Given the balanced equation:
HBr(aq) + NaOH(aq)
NaBr(ag) + H2O()
and remembering that
Molarity = moles/liter
OR
mmoles/mL
(1) (a) Calculate the number of mmoles of HBr in 100.0 mL of 0.250 MHB.
(b) Calculate the number of mmoles of NaOH in 100.0 mL of 0.250 MNAOH
©) When these two solutions are mixed the acid and base should neutralize one
another exactly. This means that all of the acid and base are completely used up;
ing reactant". Starting with the mmoles of
either the acid or base, calculate the number of mmoles of salt produced by the
either one could be considered a "li
reaction.
(d) Using the mmoles of salt produced and the total volume in mL of solution (from the
mixing of the acid and base solutions), calculate the molarity of the salt solution
produced by this reaction.
![(e) Consider this example problem:
If 100. ml of 0.100 M HCl solution is mixed with 100. mL of 0.100 M NaOH, what is
the molarity of the resulting salt solution? (assuming the volumes are additive and
ignore the change in H20, which is negligible).
HClag)
NaOH;ag)
Nacla)
H2Ou
Rxn ratio:
1 mol
1 mol
1 mol
1 mol
100 ml. x (0.100 mmol Naon
1 ml
(0.100 ттol Hнс
Mols @ Start:
100 ml x
O mol
1 ml
= 10 mmol HBr
= 10 mmol NaOH
Change
- 10 mmol
10 mmol NaOH
10 mmol
After rxn
O mmol
O mmol
10 mmol
The resulting solution contains 10 mmol NaCl in a total volume of 200 mL. The
molarity of NaCl solution is
10 mmol
= 0.05 M NaCl
200 mL
Fill in a table like this for the calculations in problem (1) (a)-(d).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa389d800-2220-480d-b079-1654ac8487de%2Fc7044113-ed79-4672-854f-f21d9eba53f2%2F942r6d_processed.jpeg&w=3840&q=75)
Transcribed Image Text:(e) Consider this example problem:
If 100. ml of 0.100 M HCl solution is mixed with 100. mL of 0.100 M NaOH, what is
the molarity of the resulting salt solution? (assuming the volumes are additive and
ignore the change in H20, which is negligible).
HClag)
NaOH;ag)
Nacla)
H2Ou
Rxn ratio:
1 mol
1 mol
1 mol
1 mol
100 ml. x (0.100 mmol Naon
1 ml
(0.100 ттol Hнс
Mols @ Start:
100 ml x
O mol
1 ml
= 10 mmol HBr
= 10 mmol NaOH
Change
- 10 mmol
10 mmol NaOH
10 mmol
After rxn
O mmol
O mmol
10 mmol
The resulting solution contains 10 mmol NaCl in a total volume of 200 mL. The
molarity of NaCl solution is
10 mmol
= 0.05 M NaCl
200 mL
Fill in a table like this for the calculations in problem (1) (a)-(d).
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