100.0 mL of HCl required 5.831 g Mg(OH)2 to titrate. The molar mass of Mg(OH)2 is 58.31 g. What is the concentration of the HCI ? The neutralization equation is 2 HCI + Mg(OH)₂ → MgCl₂ + 2 H₂O There is enough information to calculate the moles of Mg(OH)2 mol Mg(OH)2 (5.831 g Mg(OH)2 (- g Mg(OH)₂ Then use stoichiometry to convert moles of Mg(OH)2 into moles of HCI mol HCI concentration = mol Mg(OH)2 (-- a. acid i. 0.2000 p. 0.01000 d. NaOH j. 1.000 k. 0.005000 q. 0.001000 mol Mg(OH)2 We now can calculate the concentration of HCI since we have both the moles of HCI and the volume of HCI mol HCI b. base c. HCI )= LHCI r. 58.31 ) = e. NaCl I. 100.0 f. H₂O m. 0.1000 -) = s. 5.831 g. 25.00 mol Mg(OH)2 mol/L n. 0.05000 h. 0.02500 mol HCI o. 2.000
100.0 mL of HCl required 5.831 g Mg(OH)2 to titrate. The molar mass of Mg(OH)2 is 58.31 g. What is the concentration of the HCI ? The neutralization equation is 2 HCI + Mg(OH)₂ → MgCl₂ + 2 H₂O There is enough information to calculate the moles of Mg(OH)2 mol Mg(OH)2 (5.831 g Mg(OH)2 (- g Mg(OH)₂ Then use stoichiometry to convert moles of Mg(OH)2 into moles of HCI mol HCI concentration = mol Mg(OH)2 (-- a. acid i. 0.2000 p. 0.01000 d. NaOH j. 1.000 k. 0.005000 q. 0.001000 mol Mg(OH)2 We now can calculate the concentration of HCI since we have both the moles of HCI and the volume of HCI mol HCI b. base c. HCI )= LHCI r. 58.31 ) = e. NaCl I. 100.0 f. H₂O m. 0.1000 -) = s. 5.831 g. 25.00 mol Mg(OH)2 mol/L n. 0.05000 h. 0.02500 mol HCI o. 2.000
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
Section: Chapter Questions
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![100.0 mL of HCI required 5.831 g Mg(OH)2 to titrate. The molar mass of Mg(OH)2 is 58.31 g. What is the concentration of the HCI ?
The neutralization equation is
2 HCI + Mg(OH)2
MgCl₂ + 2 H₂O
There is enough information to calculate the moles of Mg(OH)2
mol Mg(OH)2
-):
(5.831 g Mg(OH)₂ (-
g Mg(OH)₂
Then use stoichiometry to convert moles of Mg(OH)2 into moles of HCI
mol HCI
concentration
a. acid
i. 0.2000
mol Mg(OH)2 (--
mol Mg(OH)2
We now can calculate the concentration of HCI since we have both the moles of HCI and the volume of HCI
mol HCI
p. 0.01000
b. base c. HCI
NaOH
j. 1.000 k. 0.005000
q. 0.001000
L HCI
r. 58.31
e. NaCl
I. 100.0
f. H₂O
m. 0.1000
s. 5.831
g. 25.00
mol Mg(OH)2
mol/L
n. 0.05000
h. 0.02500
mol HCI
o. 2.000](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F90a9e6d8-18c9-4369-a7ca-3ea93a2f0515%2F6ce3ddb6-f901-42dd-b42f-24496acb7358%2Fyojwd7g_processed.png&w=3840&q=75)
Transcribed Image Text:100.0 mL of HCI required 5.831 g Mg(OH)2 to titrate. The molar mass of Mg(OH)2 is 58.31 g. What is the concentration of the HCI ?
The neutralization equation is
2 HCI + Mg(OH)2
MgCl₂ + 2 H₂O
There is enough information to calculate the moles of Mg(OH)2
mol Mg(OH)2
-):
(5.831 g Mg(OH)₂ (-
g Mg(OH)₂
Then use stoichiometry to convert moles of Mg(OH)2 into moles of HCI
mol HCI
concentration
a. acid
i. 0.2000
mol Mg(OH)2 (--
mol Mg(OH)2
We now can calculate the concentration of HCI since we have both the moles of HCI and the volume of HCI
mol HCI
p. 0.01000
b. base c. HCI
NaOH
j. 1.000 k. 0.005000
q. 0.001000
L HCI
r. 58.31
e. NaCl
I. 100.0
f. H₂O
m. 0.1000
s. 5.831
g. 25.00
mol Mg(OH)2
mol/L
n. 0.05000
h. 0.02500
mol HCI
o. 2.000
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