Log in
Chemistry
2
answers
0
watching
501
views

2 Al(OH)3(s) + 3 H2SO4(aq) --> Al2(SO4)3(aq) + 6 H2O(l) If 0.500 mol Al(OH)3 and 0.500 mol H2SO4 react, and 48.2 g Al2(SO4)3 are collected from the experiment, what is the percent yield? The molar mass of Al2(SO4)3 is 342.17 g/mol. Be sure to include "%" units with your answer.

Watch
2
answers
0
watching
501
views

For unlimited access to Homework Help, a Homework+ subscription is required.

Unlock all answers

Get 1 free homework help answer.
Already have an account? Log in
Share
Deanna Hettinger
Deanna HettingerLv2
21 Apr 2020
Already have an account? Log in
Share
discord banner image
Join us on Discord
Chemistry Study Group
Join now

Related textbook solutions

Basic Chemistry
5 Edition,
Timberlake
ISBN: 9780134138046
Principles of Chemistry Molecular Approach
4th Edition,
Tro
ISBN: 9780134112831
Chemistry: Structure and Properties
2nd Edition,
Tro
ISBN: 9780134293936
Principles of Chemistry Molecular Approach
3rd Edition, 2014
Tro
ISBN: 9780321971944
Chemistry: A Molecular Approach
3rd Edition,
Tro
ISBN: 9780321809247
Chemistry: A Molecular Approach
5th Edition,
Tro
ISBN: 9780134874371
Principles of Chemistry: A Molecular Approach
4th Edition,
Tro
ISBN: 9780134895741
Chemistry: The Central Science
14th Edition, 2017
Brown
ISBN: 9780134414232

Related questions

Aluminum metal will reduce water, but the reaction is extremely slow under neutral conditions at room temperature.

Al(s) + 3H2O(l) → Al(OH)3(s) + H2(g) very, very slow

In the presence of aqueous base, however, Al will readily react to form the very soluble Al(OH)4–1(aq) complex anion which can be neutralized with H2SO4(aq) in two steps. The first produces the very insoluble Al(OH)3(s). If the Al(OH)3(s) slurry is then immediately boiled with additional H2SO4(aq) the insoluble hydroxide will react completely to give a clear, colorless solution. After excess water is boiled off, cooling produces KAl(SO4)2.12H2O(s), Alum, as fine, white needles.

Al(s) + 3H2O + KOH ⟶ KAl(OH)4(aq) + H2(g) fast and exothermic at room temperature

KAl(OH)4(aq) + ½ H2SO4(aq) ⟶ Al(OH)3(s) + H2O(l) + ½ K2SO4(aq)

Al(OH)3(s) + H2SO4(aq) ⟶ ½ Al2(SO4)3(aq) + 3 H2O (requires heating)

½ Al2(SO4)3(aq) + ½ K2SO4(aq) ⟶ KAl(SO4)2(s)

Combining these four equations gives the overall reaction.

Al(s) + KOH(aq) + 2 H2SO4(aq) ⟶ [KAl(SO4)2(aq)] + H2O(l) + H2(g) and

[KAl(SO4)2(aq)] + 12 H2O(l) ⟶ KAl(SO4)2.12H2O(s) (Alum)

A student dissolved 0.946g Al in 35mL of 2.00M KOH and obtained 13.8g of dry alum by following the experimental procedure given in this exercise. Assume that the Alum was 97.7% pure. Calculate the volume of 6.00 M H2SO4 (aq) needed just to precipitate Al(OH)3 (s).

Aluminum hydroxide reacts with sulfuric acid as follows:

2Al(OH)3(s)+3H2SO4(aq)→Al2(SO4)3(aq)+6H2O(l)

1) Which reagent is the limiting reactant when 0.650 mol Al(OH)3 and 0.650 mol H2SO4 are allowed to react?

2) How many moles of Al2(SO4)3 can form under these conditions? Express the amount in moles to three significant digits.

3) How many moles of the excess reactant remain after the completion of the reaction? Express the amount in moles to three significant digits.

Calculate the theoretical yield, percent yield, and atom economy of the synthesis reaction

2Al(s)+ 2KOH(aq) + 4H2SO4(aq) + 22H2O(l) =====> 2[KAl(SO4)2 * 12H2O](s) + 3H2(g)

Mass of Al is 1.000 g

Volume of KOH is 49.0 mL, KOH used is 1.5M

Mass of [KAl(SO4)2 * 12H2O](s) is .246 g ( This was experimentaly found and is not accurate, but disregard that solve the problem as if it was correct)

Molecular weight of [KAl(SO4)2 * 12H2O](s) is 474.37 grams/mole.

15 mL of 9M H2SO4(aq) is added

The following process is used

2Al(s)+ 2KOH(aq) + 6H2O(l) + heat =====> 2KAl(OH)4(aq)+ 3H2(g)

2KAl(OH)4(aq)+ H2SO4(aq) =====> 2Al(OH)3(s) + K2SO4(aq) + 2H2O(l)

2Al(OH)3 + 3H2SO4(aq) + heat =====> Al2(SO4)3(aq) + 6H2O(l)

Al2(SO4)3(aq) + K2SO4(aq) + 24H2O(l) =====> 2[KAl(SO4)2 * 12H2O](s)

Please posted questions if you are unsure about anything and I will ansewr them

Weekly leaderboard

Home
Homework Help 3,900,000
Chemistry 830,000
Start filling in the gaps now
Log in