Given the balancedequation: 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 mmoles of HBr = molarity x volume = 0.250 x 100 2o5 mmoles HBr (b) Calculate the number of mmoles of NaOH in 100.0 mL of 0.250 MNaOH mmoles of NAOH: molarity 0.250 x 100 2.5 mmoles NaOH x volueme (c) When these two solutions are mixed the acid and base shouid 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.

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3:34 PM Wed Nov 4
* 36% I
Given the balancedequation:
HBr(aq) + NAOH(aq) →
NaBr(aq) + H2OU»
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
mmoles of HBr = molarity x volume
O-250 x
100
205 mmoles HBr
(6) Calculate the number of mmoles of NaOH in 100.0 mL of 0.250 MNAOH
mmoles of NaOH = molo
Lolarity olume
0.250 x l00
%3D
2.5 mmoles NaoH
© When these two solutions are mixed the acid and pase snouid 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
base, calculate
number of mmoles of salt produced by the
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.
1
4
(e) Consider this example problem:
If 100. ml of 0.100 M HCI 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).
HClaa)
NaOHjaal
Naclja)
I mol
1 mol
1 mol
1 mol
Ryn ratio:
Transcribed Image Text:3:34 PM Wed Nov 4 * 36% I Given the balancedequation: HBr(aq) + NAOH(aq) → NaBr(aq) + H2OU» 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 mmoles of HBr = molarity x volume O-250 x 100 205 mmoles HBr (6) Calculate the number of mmoles of NaOH in 100.0 mL of 0.250 MNAOH mmoles of NaOH = molo Lolarity olume 0.250 x l00 %3D 2.5 mmoles NaoH © When these two solutions are mixed the acid and pase snouid 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 base, calculate number of mmoles of salt produced by the 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. 1 4 (e) Consider this example problem: If 100. ml of 0.100 M HCI 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). HClaa) NaOHjaal Naclja) I mol 1 mol 1 mol 1 mol Ryn ratio:
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