Check Example #5) 10mL of 3.0M HBr is mixed with 20mL of 2.0M KOHHBr +KOH → H₂O + KBrAHreaction=AHproducts-AHreactantsSubstanceH₂OHBrKBrKOHHere's the first reactant: 10mL of 3.0M HBr10* 3= 0.03 mol H Brקסיטן(-286+-392)+(-36.27 +2 480)== -161.8 kJ1000Let's calculate the limiting reagent to determine how many mols of water will be formed:*-Let's do this for the second reactant, 20mL of 2.0M KOH202 = 0.04 molk@HCalculate the amount of energy released:28636.20·03 * - 161.8=-4-854 kAHf in kJ/mol-392480 Example #6) 250mL of 2.0M HBr is mixed with 300mL of 1.5M KOHHBr + NaOH > H2O + NaBrSubstanceH₂OHBrNaBrNaOHAHreaction=AHproducts-AHreactants(-286 +-362)-(-36.2 +-42607)= -18531 kJHere's the first reactant: 250mL of 2.0M HBr250 * 21000* 2 = 0.5 mol H BrLet's calculate the limiting reagent to determine how many mols of water will be formed:1000Let's do this for the second reactant, 300mL of 1.5M KOH3002 = 0.6 mol koff*Calculate the amount of energy released:AHf in kJ/mol-28636.2-362-8-426.70.5 * - 1851 = -192.55k

Using the stoichiometry of acid to base, moles of water formed can be calculated which helps in…

Answered: Example #5) 10mL of 3.0M HBr is mixed…

Example #5) 10mL of 3.0M HBr is mixed with 20mL of 2.0M KOH HBr + KOH → H₂O + KBr Substance H₂O HBr KBr KOH AHf in kJ/mol - 286 36.2 -392 - 480

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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Question

Check

Example #5) 10mL of 3.0M HBr is mixed with 20mL of 2.0M KOH
HBr +KOH → H₂O + KBr
AHreaction=AHproducts-AHreactants
Substance
H₂O
HBr
KBr
KOH
Here's the first reactant: 10mL of 3.0M HBr
10
* 3
= 0.03 mol H Br
קסיטן
(-286+-392)+(-36.27 +2 480)
== -161.8 kJ
1000
Let's calculate the limiting reagent to determine how many mols of water will be formed:
*
-
Let's do this for the second reactant, 20mL of 2.0M KOH
20
2 = 0.04 molk@H
Calculate the amount of energy released:
286
36.2
0·03 * - 161.8=-4-854 k
AHf in kJ/mol
-392
480

Example #6) 250mL of 2.0M HBr is mixed with 300mL of 1.5M KOH
HBr + NaOH > H2O + NaBr
Substance
H₂O
HBr
NaBr
NaOH
AHreaction=AHproducts-AHreactants
(-286 +-362)-(-36.2 +-42607)
= -18531 kJ
Here's the first reactant: 250mL of 2.0M HBr
250 * 2
1000
* 2 = 0.5 mol H Br
Let's calculate the limiting reagent to determine how many mols of water will be formed:
1000
Let's do this for the second reactant, 300mL of 1.5M KOH
300
2 = 0.6 mol koff
*
Calculate the amount of energy released:
AHf in kJ/mol
-286
36.2
-362-8
-426.7
0.5 * - 1851 = -192.55k

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