Assuming you start with 1.25 g of pure aluminum, calculate the following: 1) The amount of potassium hydroxide, KOH, in grams, needed to react with all of the 1.25 g of Al. The reaction is: 2 A1 (s) + 2КОН (аq) + 6 Н20 (1) 2 K (аq) + 2 AI(ОН)4 (аq) + 3 Н2 (в) g 2) The amount of KOH needed, in grams, in reaction 1 if you want to have 10% excess KOH. g 3) The amount of water, in milliliters, that will react with the Al and KOH in reaction 1. ml
Assuming you start with 1.25 g of pure aluminum, calculate the following: 1) The amount of potassium hydroxide, KOH, in grams, needed to react with all of the 1.25 g of Al. The reaction is: 2 A1 (s) + 2КОН (аq) + 6 Н20 (1) 2 K (аq) + 2 AI(ОН)4 (аq) + 3 Н2 (в) g 2) The amount of KOH needed, in grams, in reaction 1 if you want to have 10% excess KOH. g 3) The amount of water, in milliliters, that will react with the Al and KOH in reaction 1. ml
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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![Assuming you start with 1.25 g of pure aluminum, calculate the following:
1) The amount of potassium hydroxide, KOH, in grams, needed to react with all of the 1.25 g of Al. The reaction is:
2 A1 (s) + 2КОН (аq) + 6 Н20 (1)
2 K (аq) + 2 AI(ОН)4 (аq) + 3 Н2 (в)
g
2) The amount of KOH needed, in grams, in reaction 1 if you want to have 10% excess KOH.
g
3) The amount of water, in milliliters, that will react with the Al and KOH in reaction 1.
mL
4) The amount, in milliliters, of 9 M sulfuric acid needed for reaction 2.
2 Al(OH)4 (aq) + H2SO4 (aq)
→ 2 Al(OH); (s) + SO,²- (aq) + 2 H2O (1)
mL
5) The amount, in milliliters, of 9 M sulfuric acid needed for reaction 3.
2 Al(OH)3 (s) + 3 H;SO4 (aq)
→ 2 Al³+ (aq) +3 SO,²- (aq) + 6 H2O (1)
mL](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc26804e2-ad2a-4620-909b-cc58fe508113%2F8a88431e-bfdf-4285-a5ca-9d5593695aae%2F4j4nuti_processed.png&w=3840&q=75)
Transcribed Image Text:Assuming you start with 1.25 g of pure aluminum, calculate the following:
1) The amount of potassium hydroxide, KOH, in grams, needed to react with all of the 1.25 g of Al. The reaction is:
2 A1 (s) + 2КОН (аq) + 6 Н20 (1)
2 K (аq) + 2 AI(ОН)4 (аq) + 3 Н2 (в)
g
2) The amount of KOH needed, in grams, in reaction 1 if you want to have 10% excess KOH.
g
3) The amount of water, in milliliters, that will react with the Al and KOH in reaction 1.
mL
4) The amount, in milliliters, of 9 M sulfuric acid needed for reaction 2.
2 Al(OH)4 (aq) + H2SO4 (aq)
→ 2 Al(OH); (s) + SO,²- (aq) + 2 H2O (1)
mL
5) The amount, in milliliters, of 9 M sulfuric acid needed for reaction 3.
2 Al(OH)3 (s) + 3 H;SO4 (aq)
→ 2 Al³+ (aq) +3 SO,²- (aq) + 6 H2O (1)
mL
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