All categories

3 years ago

Fe2O3 +3CO--2Fe+3CO2 .identify the substance oxidized ,reduced and oxidizing agent and reducing agent

Chemistry

1 answer:

VikaD [51]3 years ago

6 0

Answer:

Oxidized : CO

Reduced: Fe

Oxidizing agent: Fe2O3

Reducing agent: CO

Explanation:

Loss of electrons is oxidation while gaining of electrons is reduction.

CO is a reducing agent, Fe2O3 is an oxidizing agent.

Each carbon atom is oxidized in CO. CO is the reducing agent

Each Fe atom in FeO3 is reduced. FeO3 is the oxidizing agent

You might be interested in

Determine the empirical formula for a compound that contains c, h and o. it contains 52.14% c and 34.73% o by mass.

Travka [436]

H %= 100- (52.14 + 34. 73) equals 13.13 %
Assuming 100 g of this compound
Mass H= 13.13 g
Moles H= 13.13 g ÷ 1.008g/ moles= 13

Mass C= 52.14 g
Moles C= 52.14 g ÷ 12.011 g/ moles= 4

The empirical formula is C4H1302

3 0

4 years ago

Read 2 more answers

2. Identify items in your life that are made of each type of element: metal, nonmetal, and metalloid. For each item, explain why

Bad White [126]

Metalloids:alloys of metallods are commonly used in discs like compact discs, CDs ,and DVDs .These materials change from glassy to crystalline with the application of heat which makes them useful as data storage media

3 0

3 years ago

What is the formula and name of the compound produced at the end of the virtual lab, and what caused it to have a different appe

Iteru [2.4K]

Answer: $SnO_2$ , $NO_2$ and $H_2O$ are formed at the end of the reaction. They are named as tin (IV) oxide or stannic oxide, nitrogen dioxide and water respectively.

Explanation: Reaction of tin and nitric acid is given as:

$Sn(s)+4HNO_3(aq.)\rightarrow SnO_2(s)+4NO_2(g)+2H_2O(l)$

Three products are formed at the end of the reaction which are:

$SnO_2$ which is termed as stannic oxide or Tin (IV) oxide. This is a white colored solid.

$NO_2$ which is termed as nitrogen dioxide. These are brown colored fumes.

$H_2O$ which is termed as water.

At the starting tin was a silvery-white colored solid and after the reaction, it changed the color to milky-white. This change in color is due to the chemical reaction happening between tin and nitric acid.

Release of brown fumes are also an indication that a chemical reaction has taken place.

6 0

3 years ago

Escribe la configuración electrónica de los elementos que poseen, ¿de qué elementos se trata?: 15 electrones 27 electrones 56 el

allsm [11]

Answer:

15 electrones: 1S²2S²2P⁶3S²3P³. Fósforo

27 electrones: 1S²2S²2P⁶3S²3P⁶4S²3d⁷ - Cobalto.

56 electrones: 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s² - Bario

49 electrones: 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p¹ - Indio

Explanation:

Para llenar los orbitales electrónicos de los distintos átomos debemos hacer uso de la regla de llenado electrónico de Aufbau. Por ejemplo, para el átomo con 15 electrones, la configuración electrónica es:

1S²2S²2P⁶3S²3P³. 2+2+6+2+3 = 15 electrones

Si elemento es neutro, tiene 15 protones. Es decir, es el fósforo, P.

27 electrones:

1S²2S²2P⁶3S²3P⁶4S²3d⁷ - Cobalto.

56 electrones:

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s² - Bario

49 electrones:

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p¹ - Indio

5 0

3 years ago

If 21.42g of KMnO4 is actually produced what is the percent yield

mrs_skeptik [129]

Answer:

Percentage yield = 6.776%

Explanation:

Data Given:

Actual yield of KMnO₄ = 21.42g

Percentage Yield = ?

Formula Used to find Percent yield

Percentage yield = Actual yield/ theoretical yield x 100 (1)

For this Pupose First step is

We have to know the theoretical yield KMnO₄

Potasium permagnate form from MnO₄ and KOH in the presence of Oxygen by heating, in 1st step in second 2K₂MnO₄ react with HCl and give KMnO₄ .

The Reactions of formation of KMnO₄

1st Step

2MnO₄ + 4KOH + O₂ ------------> 2K₂MnO₄ + 2H₂O

2nd Step

2K₂MnO₄ + 4HCl ------------->2 KMnO₄ + H₂O + 4KCl

So form the above equation we come to know it produced 2 Mole of KMnO₄

Now we will calculate the mass of KMnO₄ by mass formulae

mass of KMnO₄ = Number of mole of KMnO₄ x Molar Mass of KMnO₄

Molar Mass of KMnO₄ = 158.034g /mol

Put value in Mass Formula

mass of KMnO₄ = 2 mol x 158.034g/mol

mass of KMnO₄ = 316.1 g

So the theoratical yield per standard reaction = 316.1 g

Now put all values in equation 1

Percentage yield = Actual yield/ theoretical yield x 100

Percentage yield = 21.42g / 316.1 g x 100

Percentage yield = 0.0678 x 100

Percentage yield = 6.776%

4 0

3 years ago

Fe2O3 +3CO--2Fe+3CO2 .identify the substance oxidized ,reduced and oxidizing agent and reducing agent​

Fe2O3 +3CO--2Fe+3CO2 .identify the substance oxidized ,reduced and oxidizing agent and reducing agent