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Xelga [282]
2 years ago
11

Which element or compound has lost electrons in this oxidation-reduction reaction? 4li 2coo → 2co 2li2o a. coo b. li2o c. li d

. co e. o
Chemistry
1 answer:
Nady [450]2 years ago
7 0

Answer:

       Correct option is C.

Explanation:

Such type of chemical reaction that involves a transfer of electrons between two species is known as oxidation-reduction or redox reaction.

Redox reaction is oxidation number of a molecule, atom, or ion changes by gaining or losing an electron.

For this reaction:

           On reactant side:

Oxidation state of oxygen, cobalt, lithium = -2 ,+2, 0

While on product side of reaction:

Oxidation state of cobalt, oxygen, lithium = 0, -2, +1

Here, oxidation reaction take place because oxidation state of lithium is increased. While the oxidation state of cobalt in reducing, it is gaining electrons and so it undergoes reduction reaction.

So, Option C (Li) is correct.

Learn more about oxidation-reduction reaction:

brainly.com/question/13293425

#SPJ4

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<u>Answer:</u> The mass of nickel (II) oxide and aluminium that must be used is 18.8 g and 4.54 g respectively.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}      .....(1)

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  • <u>For nickel (II) oxide:</u>

By Stoichiometry of the reaction:

3 moles of nickel are produced from 3 moles of nickel (II) oxide

So, 0.252 moles of nickel will be produced from \frac{3}{3}\times 0.252=0.252mol of nickel (II) oxide

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Moles of nickel (II) oxide = 0.252 moles

Putting values in equation 1, we get:

0.252mol=\frac{\text{Mass of nickel (II) oxide}}{74.7g/mol}\\\\\text{Mass of nickel (II) oxide}=(0.252mol\times 74.7g/mol)=18.8g

  • <u>For aluminium:</u>

By Stoichiometry of the reaction:

3 moles of nickel are produced from 2 moles of aluminium

So, 0.252 moles of nickel will be produced from \frac{2}{3}\times 0.252=0.168mol of aluminium

Now, calculating the mass of aluminium by using equation 1:

Molar mass of aluminium = 27 g/mol

Moles of aluminium = 0.168 moles

Putting values in equation 1, we get:

0.168mol=\frac{\text{Mass of aluminium}}{27g/mol}\\\\\text{Mass of aluminium}=(0.168mol\times 27g/mol)=4.54g

Hence, the mass of nickel (II) oxide and aluminium that must be used is 18.8 g and 4.54 g respectively.

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