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QveST [7]
3 years ago
7

Identify the oxidizing and reducing agents in this reaction: Cr2O3(s) + 2Al(s) 2Cr(s) + Al2O3(s)

Chemistry
2 answers:
DiKsa [7]3 years ago
7 0

In this reaction, Cr2O3 is the oxidizing agent which will oxidize Al to Al2O3.

Al is the reducing agent that will reduce Cr2O3 to Cr.

An oxidizing agent has the property of oxidizing another molecule (giving it one or more oxygen atoms) by being reduced.

A reducing agent has the property of reducing another molecule (by removing one or more oxygen molecules) by being oxidized.

aivan3 [116]3 years ago
3 0
Cr2O3+ 2Al -----------> 2Cr+ Al2O3

Cr2O3--------2Cr.
( oxidation no. of cr +3 to +1 means hain of electron so it act as oxidizing agent)

similarly,

2Al------> Al2O3
( oxidation no. of Al +1 to +3 means loss of electron so it act as reducing agent)
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6 0
3 years ago
Ba(oh)2+H3po4+h2o how is it <br> balance ?
Igoryamba

Answer:

3Ba(OH)2 + 2H3PO4 —> Ba3(PO4)2 + 6H2O

Explanation:

Ba(OH)2 + H3PO4 —> Ba3(PO4)2 + H2O

There are 3 atoms of Ba on the right side and 1atom on the left side. It can be balance by putting 3 in front of Ba(OH)2 as shown below:

3Ba(OH)2 + H3PO4 —> Ba3(PO4)2 + H2O

There are 2 atoms of P on the right side and 1atom on the left. It can be balance by putting 2 in front of H3PO4 as shown below:

3Ba(OH)2 + 2H3PO4 —> Ba3(PO4)2 + H2O

Now, there are a total of 12 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 6 in front of H2O as shown below:

3Ba(OH)2 + 2H3PO4 —> Ba3(PO4)2 + 6H2O

Now the equation is balanced as the numbers of the atoms of the different elements present on both sides are equal

4 0
3 years ago
How much heat is released when 75 g of octane is burned completely if the enthalpy of combustion is -5,500 kJ/mol CsH18? C8H18 +
aalyn [17]

Answer : The correct option is, (D) 3600 kJ

Explanation :

Mass of octane = 75 g

Molar mass of octane = 114.23 g/mole

Enthalpy of combustion = -5500 kJ/mol

First we have to calculate the moles of octane.

\text{ Moles of octane}=\frac{\text{ Mass of octane}}{\text{ Molar mass of octane}}=\frac{75g}{114.23g/mole}=0.656moles

Now we have to calculate the heat released in the reaction.

As, 1 mole of octane released heat = -5500 kJ

So, 0.656 mole of octane released heat = 0.656 × (-5500 kJ)

                                                                   = -3608 kJ

                                                                   ≈ -3600 kJ

Therefore, the heat released in the reaction is 3600 kJ

5 0
3 years ago
AnonymousXDX can you answer this question to
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Answer:

.

Explanation:

3 0
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