How to get fe oh 3 from fe2o3. Iron and its compounds
The human body contains about 5 g of iron, most of it (70%) is part of blood hemoglobin.
Physical properties
In its free state, iron is a silvery-white metal with a grayish tint. Pure iron is ductile and has ferromagnetic properties. In practice, iron alloys - cast iron and steel - are usually used.
Fe is the most important and most abundant element of the nine d-metals of the Group VIII subgroup. Together with cobalt and nickel it forms the “iron family”.
When forming compounds with other elements, it often uses 2 or 3 electrons (B = II, III).
Iron, like almost all d-elements of group VIII, does not exhibit a higher valency equal to the group number. Its maximum valency reaches VI and appears extremely rarely.
The most typical compounds are those in which the Fe atoms are in oxidation states +2 and +3.
Methods for obtaining iron
1. Technical iron (alloyed with carbon and other impurities) is obtained by carbothermic reduction of its natural compounds according to the following scheme:
Recovery occurs gradually, in 3 stages:
1) 3Fe 2 O 3 + CO = 2Fe 3 O 4 + CO 2
2) Fe 3 O 4 + CO = 3FeO + CO 2
3) FeO + CO = Fe + CO 2
The cast iron resulting from this process contains more than 2% carbon. Subsequently, cast iron is used to produce steel - iron alloys containing less than 1.5% carbon.
2. Very pure iron is obtained in one of the following ways:
a) decomposition of Fe pentacarbonyl
Fe(CO) 5 = Fe + 5СО
b) reduction of pure FeO with hydrogen
FeO + H 2 = Fe + H 2 O
c) electrolysis of aqueous solutions of Fe +2 salts
FeC 2 O 4 = Fe + 2CO 2
iron(II) oxalate
Chemical properties
Fe is a metal of medium activity and exhibits general properties characteristic of metals.
A unique feature is the ability to “rust” in humid air:
In the absence of moisture with dry air, iron begins to react noticeably only at T > 150°C; upon calcination, “iron scale” Fe 3 O 4 is formed:
3Fe + 2O 2 = Fe 3 O 4
Iron does not dissolve in water in the absence of oxygen. At very high temperatures, Fe reacts with water vapor, displacing hydrogen from water molecules:
3 Fe + 4H 2 O(g) = 4H 2
The mechanism of rusting is electrochemical corrosion. The rust product is presented in a simplified form. In fact, a loose layer of a mixture of oxides and hydroxides of variable composition is formed. Unlike the Al 2 O 3 film, this layer does not protect iron from further destruction.
Types of corrosion
Protecting iron from corrosion
1. Interaction with halogens and sulfur at high temperatures.
2Fe + 3Cl 2 = 2FeCl 3
2Fe + 3F 2 = 2FeF 3
Fe + I 2 = FeI 2
Compounds are formed in which the ionic type of bond predominates.
2. Interaction with phosphorus, carbon, silicon (iron does not directly combine with N2 and H2, but dissolves them).
Fe + P = Fe x P y
Fe + C = Fe x C y
Fe + Si = Fe x Si y
Substances of variable composition are formed, such as berthollides (the covalent nature of the bond predominates in the compounds)
3. Interaction with “non-oxidizing” acids (HCl, H 2 SO 4 dil.)
Fe 0 + 2H + → Fe 2+ + H 2
Since Fe is located in the activity series to the left of hydrogen (E° Fe/Fe 2+ = -0.44 V), it is capable of displacing H 2 from ordinary acids.
Fe + 2HCl = FeCl 2 + H 2
Fe + H 2 SO 4 = FeSO 4 + H 2
4. Interaction with “oxidizing” acids (HNO 3, H 2 SO 4 conc.)
Fe 0 - 3e - → Fe 3+
Concentrated HNO 3 and H 2 SO 4 “passivate” iron, so at ordinary temperatures the metal does not dissolve in them. With strong heating, slow dissolution occurs (without releasing H 2).
In the section HNO 3 iron dissolves, goes into solution in the form of Fe 3+ cations and the acid anion is reduced to NO*:
Fe + 4HNO 3 = Fe(NO 3) 3 + NO + 2H 2 O
Very soluble in a mixture of HCl and HNO 3
5. Relation to alkalis
Fe does not dissolve in aqueous solutions of alkalis. It reacts with molten alkalis only at very high temperatures.
6. Interaction with salts of less active metals
Fe + CuSO 4 = FeSO 4 + Cu
Fe 0 + Cu 2+ = Fe 2+ + Cu 0
7. Reaction with gaseous carbon monoxide (t = 200°C, P)
Fe (powder) + 5CO (g) = Fe 0 (CO) 5 iron pentacarbonyl
Fe(III) compounds
Fe 2 O 3 - iron (III) oxide.
Red-brown powder, n. r. in H 2 O. In nature - “red iron ore”.
Methods of obtaining:
1) decomposition of iron (III) hydroxide
2Fe(OH) 3 = Fe 2 O 3 + 3H 2 O
2) pyrite firing
4FeS 2 + 11O 2 = 8SO 2 + 2Fe 2 O 3
3) nitrate decomposition
Chemical properties
Fe 2 O 3 is a basic oxide with signs of amphotericity.
I. The main properties are manifested in the ability to react with acids:
Fe 2 O 3 + 6H + = 2Fe 3+ + ZN 2 O
Fe 2 O 3 + 6HCI = 2FeCI 3 + 3H 2 O
Fe 2 O 3 + 6HNO 3 = 2Fe(NO 3) 3 + 3H 2 O
II. Weak acid properties. Fe 2 O 3 does not dissolve in aqueous solutions of alkalis, but when fused with solid oxides, alkalis and carbonates, ferrites form:
Fe 2 O 3 + CaO = Ca(FeO 2) 2
Fe 2 O 3 + 2NaOH = 2NaFeO 2 + H 2 O
Fe 2 O 3 + MgCO 3 = Mg(FeO 2) 2 + CO 2
III. Fe 2 O 3 - feedstock for the production of iron in metallurgy:
Fe 2 O 3 + ZS = 2Fe + ZSO or Fe 2 O 3 + ZSO = 2Fe + ZSO 2
Fe(OH) 3 - iron (III) hydroxide
Methods of obtaining:
Obtained by the action of alkalis on soluble Fe 3+ salts:
FeCl 3 + 3NaOH = Fe(OH) 3 + 3NaCl
At the time of preparation, Fe(OH) 3 is a red-brown mucous-amorphous sediment.
Fe(III) hydroxide is also formed during the oxidation of Fe and Fe(OH) 2 in moist air:
4Fe + 6H 2 O + 3O 2 = 4Fe(OH) 3
4Fe(OH) 2 + 2H 2 O + O 2 = 4Fe(OH) 3
Fe(III) hydroxide is the end product of the hydrolysis of Fe 3+ salts.
Chemical properties
Fe(OH) 3 is a very weak base (much weaker than Fe(OH) 2). Shows noticeable acidic properties. Thus, Fe(OH) 3 has an amphoteric character:
1) reactions with acids occur easily:
2) fresh precipitate of Fe(OH) 3 dissolves in hot conc. solutions of KOH or NaOH with the formation of hydroxo complexes:
Fe(OH) 3 + 3KOH = K 3
In an alkaline solution, Fe(OH) 3 can be oxidized to ferrates (salts of iron acid H 2 FeO 4 not released in the free state):
2Fe(OH) 3 + 10KOH + 3Br 2 = 2K 2 FeO 4 + 6KBr + 8H 2 O
Fe 3+ salts
The most practically important are: Fe 2 (SO 4) 3, FeCl 3, Fe(NO 3) 3, Fe(SCN) 3, K 3 4 - yellow blood salt = Fe 4 3 Prussian blue (dark blue precipitate)
b) Fe 3+ + 3SCN - = Fe(SCN) 3 thiocyanate Fe(III) (blood red solution)
Iron oxide III is a compound of oxygen and iron and is an inorganic substance. Formula Fe2O3.
Physical properties:
- solid state
- red-brown color
- thermally stable
- melting point 1566 °C
- density 5.242 g/cm3
Chemical properties:
- does not react with water
- fuses with oxides of other metals and forms double oxides - spinels
- reacts slowly with alkalis and acids
Application:
- polishing agent for glass and steel
- production of colored mineral paints and cement
- raw materials for iron smelting
- thermite welding
- storage medium (digital and analogue) on magnetic tapes
- catalyst for ammonia production
- ceramics production
- food industry (E172)
Preparation of iron oxide 3
Method 1. In a 400-600 ml glass, pour 50 ml of nitric acid (HNO3) and a little water. Next, add iron little by little.
When all the iron has dissolved, it is necessary to filter the liquid from various impurities. After filtration, a red liquid should remain. Add a solution of potassium hydroxide (KOH) to it.
In the solution, a precipitate immediately begins to form (that’s what we need). Filter the solution. Place the collected precipitate (Fe(OH)3) on an iron or steel plate (foil cannot be used) and place it in an oven heated to 100 degrees.
The output is the following powder (Fe2O3):
Method 2. Add a little hydrogen peroxide (H2O2) to a glass of hydrochloric acid (HCl). Next, add iron to the solution. A reaction will begin, during which you need to gradually add hydrogen peroxide.
The solution will begin to turn yellow and then dark red.
Then add a small amount of water and potassium hydroxide. A black precipitate (Fe(OH)) begins to form, which turns brown in air.
And we send the sediment into an oven heated to 700 °C.
Method 3. Thoroughly mix 100 g of iron sulfate (FeSO4) and 50 g of soda ash (Na2CO3). Place in a frying pan and place on high heat. Heat the mixture, stirring occasionally. As the powder heats up, it will change colors (blue -> dark purple -> black -> auburn). When the color of the powder turns red, increase the heat and heat for about 20 minutes, remembering to stir. After time has passed, remove from heat and cool the mixture (Fe2O3).
Based on materials from the site: pirotehnika.ruhelp.com
The main methods for producing sodium hydroxide are the interaction of sodium with water, the interaction of soda with slaked lime and electrolysis of an aqueous solution of table salt. Let's write these equations:
Reaction of sodium with water
2Na + 2H 2 O →2NaOH + H 2
Interaction of soda with slaked lime
Na 2 CO 3 + Ca(OH) 2 →2NaOH + CaCO 3 ↓
Electrolysis of an aqueous solution of table salt
2NaCl + 2H 2 O (electrolysis) → 2NaOH + H 2 + Cl 2
Iron II hydroxide is an insoluble base, therefore it can easily be obtained by the interaction of soluble iron II acid and any alkali, in addition, it can be obtained by the interaction of the oxide and water. All reactions must be carried out without access to air, since in air iron II hydroxide quickly oxidizes into iron III hydroxide. ( 4 Fe(OH) 2 +2H 2 O+O 2 =4 Fe(OH) 3 ). Let's write the equations:
FeSO 4 + 2NaOH → Fe(OH) 2 ↓ + Na 2 SO 4
FeCl 2 + 2KOH → Fe(OH) 2 ↓ + 2KCl
FeO + H 2 O → Fe(OH) 2 ↓
Without making calculations, calculate the sign of the entropy change for the reactions:
2CH 4 (g) ↔ C 2 H 2 (g) + 3H 2 (g)
N 2(g) + 3H 2(g) ↔ 2NH 3(g)
2C (graphite) + O 2(g) ↔ 2CO (G)
In case:
a) from 2 moles of gaseous starting substances we get 4 moles of gaseous products, therefore, the entropy will increase, hence the sign of the change in entropy is “+”.
b) from 4 moles of gaseous starting substances we get 2 moles of gaseous products, therefore, the entropy will decrease, therefore the sign of the change in entropy is “–”
c) from 1 mole of gaseous starting substances we get 2 moles of gaseous products, therefore, the entropy will increase, hence the sign of the change in entropy is “+”.
^ How many grams of a metal, the equivalent molar mass of which is 12.15 g/mol, reacts with 112 cm 3 of oxygen under standard conditions?
The heterogeneous reaction C (k) + CO 2 (g) ↔ 2CO (g) determines the course of all processes of carbothermic production of metals from oxides. How many times will the rate of this reaction change when the pressure of the system decreases by a factor of four? Confirm your answer with calculations.
According to the law of mass action, the rate of a chemical reaction is directly proportional to the concentration of reactants taken to degrees equal to their stoichiometric coefficients. We need to find how the rate of the forward reaction will change. Since the reaction is heterogeneous, the rate of the chemical reaction will depend only on the concentration of the gaseous phase, that is, on the concentration of carbon dioxide, therefore, the mathematical expression of the law of mass action for this reaction will be:
v=k[CO 2 ]
Let at the initial moment [CO 2 ] (init) = x, then after reducing the system pressure by 4 times, the concentration of carbon dioxide will also decrease by 4 times, that is, [CO 2 ] (con) = 0.25x
Hence:
v 1 = k[WITHO 2 ] (beginning) = kx;
v 2 =k[CO 2 ] (kon) =k0.25x
As can be seen from the calculations, the reaction rate before the pressure change is 4 times greater, therefore, reducing the system pressure by 4 times will lead to a decrease in the rate of the direct reaction by 4 times.
Answer: will decrease by 4 times
Reduction of WCl 6 vapors with hydrogen is one of the methods for producing tungsten WCl 6 (g) + 3H 2 (g) ↔ W (k) + 6HCl (g), ∆ r H 0 = 44.91 kJ. How should pressure and temperature be changed to increase metal yield?
We need to increase the yield of metal, therefore we need to shift the equilibrium towards the reaction products (to the right).
Since we get 6 moles of gaseous products from 4 moles of gaseous products, therefore, during the direct process, the pressure in the system increases, therefore, to shift the equilibrium to the right, according to LeChatelier’s principle, we need to lower the pressure.
Since the direct reaction occurs with the absorption of heat, to shift the equilibrium to the right we need to increase the temperature.
To increase the metal yield, it is necessary to lower the pressure and increase the temperature.
^ Determine the molar concentration of the equivalent solution if 0.1 mol of KOH is dissolved in 200 ml?
^ For the molecular equation Na 2 SO 3 + 2HCl ↔ 2NaCl + H 2 SO 3, write the ionic-molecular equation.
molecular:
Na 2 SO 3 + 2HCl ↔ 2NaCl + H 2 SO 3
complete ion-molecular:
2Na + + SO 3 2- + 2H + + 2Cl - → 2Na + + 2Cl - +H2SO3
abbreviated ion-molecular:
2H + + SO 3 2- → H 2 SO 3
Write molecular and ionic-molecular equations for the hydrolysis of salts: CaCO 3, ZnSO 4, (NH 4) 2 S. Specify the solution medium. Where will the hydrolysis equilibrium shift when alkali is added to a solution of each salt?
I stage:
2CaCO 3 + 2HOH → Ca(HCO 3) 2 + Ca(OH) 2
2Ca 2+ + 2CO 3 2- + 2HOH →Ca 2+ + 2HCO 3 - + Ca 2+ + 2OH -
CO 3 2- + HOH →HCO 3 - + OH -
II stage:
Ca(HCO 3) 2 + 2HOH → Ca(OH) 2 + 2H 2 CO 3
Ca 2+ + 2HCO 3 - + 2HOH → Ca 2+ + 2OH - + 2H 2 CO 3
HCO 3 - + HOH → H 2 CO 3 + OH -
› therefore pH›7 (alkaline)
The addition of alkali will increase the concentration of hydroxide ions in the solution, that is, the concentration of the product of the reversible reaction will increase, therefore, according to LeChatelier’s principle, the equilibrium of hydrolysis will shift towards the starting substances (to the left).
Hydrolysis ZnSO 4 ( a salt of a strong acid and a weak base, therefore hydrolysis will occur along the cation)
I stage:
2ZnSO 4 + 2HOH →(ZnOH) 2 SO 4 + H 2 SO 4
2Zn 2+ + 2SO 4 2- + 2HOH →2ZnOH + + SO 4 2- + 2H + + SO 4 2-
Zn 2+- + HOH →ZnOH + + H +
II stage:
(ZnOH) 2 SO 4 + 2HOH → 2Zn(OH) 2 + H 2 SO 4
2ZnOH + + SO 4 2 + 2HOH→ 2Zn(OH) 2 + 2H + + SO 4 2-
^ ZnOH + + HOH → Zn(OH) 2 + H +
‹ therefore pH‹7 (acidic)
The addition of alkali will increase the concentration of hydroxide ions in the solution, which will bind the hydride ions formed as a result of hydrolysis, that is, the concentration of the reversible reaction product will decrease, therefore, according to LeChatelier’s principle, the hydrolysis equilibrium will shift towards the hydrolysis products (to the right).
Hydrolysis (N.H. 4 ) 2 S ( a salt of a weak acid and a weak base, therefore hydrolysis will proceed along the cation and the anion)
(NH 4) 2 S + 2HOH → 2NH 4 OH + H 2 S
2NH 4 + + S 2- + 2HOH →2NH 4 OH + H 2 S
= therefore pH=7 (neutral environment)
In this case, adding alkali will not affect the chemical equilibrium of ammonium sulfide hydrolysis.
Complete the reaction equation and arrange the coefficients using the electron-ion method
