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13 Dec 2019

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).

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Jamar Ferry
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17 Dec 2019

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Related questions

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

DATA FROM EXPERIMENT:

MASS OF WEIGHING PAPER = 0.3670 g
MASS OF WEIGHING PAPER + Al FOIL = 1.0000 g
MASS OF ALUM FOIL (NET WEIGHT OF ALUM) = 0.6330 g
MASS OF EMPTY 50 mL BEAKER = 29.3380 g
MASS OF BEAKER + ALUM CRYSTALS = 39.8840 g
MASS OF ALUM (NET WEIGHT OF ALUM) = 10.5460 g

we added about 35 mL of de-ionized water and 12mL of 6M KOHsolution to the beaker. I converted the molarity to moles and got6M KOH (.012 L) = .072 mol KOH

When metallic aluminum comes into contact with aqueous solutions ofstrong bases like potassium hydroxide, KOH, it reacts to formhydrogen gas and a salt containing both aluminum and potassiumions.

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

This salt, KAl(OH)4, is soluble in solutions of strong acids, suchas sulfuric acid, H2 SO4, and reacts to form aluminum hydroxide,Al(OH)3, and potassium sulfate, K2 SO4.

2 KAl(OH)4 (aq) + H2 SO4 (aq) ? 2 Al(OH)3 (s) + K2 SO4 (aq) + 2 H2O(l)

Excess sulfuric acid is required to dissolve the Al(OH)3, and toform aluminum sulfate, Al2 (SO4)3.

2 Al(OH)3 (s) + 3 H2 SO4 (aq) ? Al2 (SO4)3 (aq) + 6 H2O (l)

As this solution is cooled, the aluminum and potassium sulfatesalts crystallize out of the solution as the alum. The equationbelow shows only the ions and the water involved in thecrystallization process.

K+ (aq) + Al3+ (aq) + 2 SO4 2- (aq) + 12 H2O (l) ? KAl(SO4)2 * 12H2O (s)

MY QUESTION IS.. I DONT KNOW WHICH EQUATION TO USE TO CALCULATE THEPERCENT YIELD. SHOULD IT BE THE FIRST ONE SINCE IT IS BEFORE ANY OFTHE PROCEDURE WAS DONE? BUT FROM THE FIRST EQUATION, IT DID NOTPRODUCE THE ALUM CRYSTALS YET : KAl(SO4)2 * 12H20

EQUATION FOR PERCENT YIELD:
100 % x [mass of actual yield / mass of theoretical yield]

Balance the following Chemical Equations:

____Fe(s) + ____H2O(g) ® ____Fe3O4(s) + ____H2(g)

____C2H6(g) + ____O2(g) ® ____CO2(g) + ____H2O(g)

(C) ___ KClO3 (s) ® ___ KCl (s) + ___ O2 (g)

___ C6H14 (l) + ___ O2 (g) ® ___ CO2 (g) + ___ H2O (g)

___ C3H8O3 (l) + ___ O2 (g) ® ___ CO2 (g) + ___ H2O (l)

___ NaN3 (s) ® ___ Na (s) + ___ N2 (g)

__ C3H5N3O9 ® __ CO2 + ___ N2 + ___ H2O + ___ O2

Write and Balance the equations:

Aqueous calcium hydroxide reacts with aqueous sodium carbonate to produce solid calcium carbonate and aqueous sodium hydroxide.

Solid iron reacts with Oxygen gas to form solid Iron (II) oxide

Copper(II) sulfide is oxidized by molecular oxygen to produce gaseous sulfur trioxide and solid copper(II) oxide. The gaseous product then reacts with liquid water to produce liquid hydrogen sulfate as the only product. Write the two equations which represent these reactions.

Indicate what type, or types, of reaction each of the following represents: (precipitation, acid-base, and oxidation-reduction.)

(A) H2O(g) + C(s) ⟶ CO(g) +H2(g)

(B) 2KClO3(s) ⟶ 2KCl(s) +3O2(g)

(C) Al(OH)3(aq) +3HCl(aq) ⟶ AlCl3(aq) + 3H2O(l)

(D) Pb(NO3)2(aq) + H2SO4(aq) ⟶ PbSO4(s) + 2HNO3(aq)

4. Silver can be separated from gold because silver dissolves in nitric acid while gold does not. Is the dissolution of silver in nitric acid an acid-base reaction or an oxidation-reduction reaction? Explain your answer.

5. Determine the oxidation states of the elements in the compounds listed. None of the oxygen-containing compounds are peroxides or superoxides.

(A) H3PO4

(B) Al(OH)3

(C) SeO2

6. Complete and balance the following acid-base equations:

(A) A solution of HClO4 is added to a solution of LiOH.

(B) Aqueous H2SO4 reacts with NaOH.

7. Write the molecular, total ionic, and net ionic equations for the following reactions:

(A) Ca(OH)2(aq) + HC2H3O2(aq) ⟶

(B) H3PO4(aq) + CaCl2(aq) ⟶

8. Write the balanced equation, then calculate the number of moles and the mass of Mg required to react with 5.00 g of HCl and produce MgCl2 and H2. Ans: 1.67 g Mg

9. H2 is produced by the reaction of 118.5 mL of a 0.8775-M solution of H3PO4 according to the following equation: 2Cr + 2H3PO4 ⟶ 3H2 + 2CrPO4. How many grams of H2 are produced? Ans: .315 g H2

10. Automotive air bags inflate when a sample of sodium azide, NaN3, is very rapidly decomposed.

2NaN3(s) ⟶ 2Na(s) + 3N2(g)

What mass of sodium azide is required to produce 2.6 ft3 (73.6 L) of nitrogen gas with a density of 1.25 g/L?

Ans: 142 g NaN3

11. Urea, CO(NH2)2, is manufactured on a large scale for use in producing urea-formaldehyde plastics and as a fertilizer. What is the maximum mass of urea that can be manufactured from the CO2 produced by combustion of 1.00 x 104 grams of CO2?

CO2(g) + 2NH3(g) ⟶ CO(NH2)2(s) + H2O(l)

Ans: 1360 g urea

12. What volume o f 0.750 M hydrochloric acid solution can be prepared from the HCl produced by the reaction of 25.0 g of NaCl with excess sulfuric acid? NaCl(s) + H2SO4(l) ⟶ HCl(g) + NaHSO4(s)

Ans: .570 L

13. A sample of 0.53 g of carbon dioxide was obtained by heating 1.31 g of calcium carbonate. What is the percent yield for this reaction? CaCO3(s)⟶CaO(s) + CO2(s)

Ans: 92.0 %

14. Citric acid, C6H8O7, a component of jams, jellies, and fruity soft drinks, is prepared industrially via fermentation of sucrose by the mold Aspergillus niger. The equation representing this reaction is

C12H22O11 + H2O +3O2 ⟶2C6H8O 7+ 4H2O

What mass of citric acid is produced from the reaction of 150. grams of sucrose if the yield is 92.30%?

Ans: 155 g

15. The phosphorus pentoxide used to produce phosphoric acid for cola soft drinks is prepared by burning phosphorus in oxygen.

(A) What is the limiting reactant when 0.200 mol of P4 and 0.200 mol of O2 react according to P4+5O2 ⟶ P4O10?

(B) Calculate the percent yield if 10.0 g of P4O10 is isolated from the reaction.

Ans: 88.0 %

16. What is the limiting reactant when 1.50 g of lithium and 1.50 g of nitrogen combine to form lithium nitride, a component of advanced batteries, according to the following unbalanced equation? Li + N2 ⟶ Li3N

Ans: Li

17. What is the concentration of NaCl in a solution if titration of 15.00 mL of the solution with 0.2503 M AgNO3 requires 20.22 mL of the AgNO3 solution to reach the end point?

AgNO3(aq) + NaCl(aq) ⟶ AgCl(s) +NaNO (aq)

Ans: .3374 M

18. In a common medical laboratory determination of the concentration of free chloride ion in blood serum, a serum sample is titrated with a Hg(NO3)2 solution.

2Cl−(aq) + Hg(NO3)2(aq) ⟶ 2NO3−(aq) + HgCl2(s)

What is the Cl− concentration in a 0.25-mL sample of normal serum that requires 1.46 mL of .000825 M Hg(NO3)2(aq) to reach the end point?

Ans: 0.009636 M

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