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Elena L [17]
3 years ago
14

5. Given the reaction:

Chemistry
1 answer:
Alja [10]3 years ago
6 0

Answer:

48 g of oxygen is needed to produced 54 g of water.

Explanation:

Given data:

Mass of O₂ = ?

Mass of H₂O = 54 g

Chemical equation:

O₂  + 2H₂  → 2 H₂O

Number of moles of H₂O:

Number of moles of  H₂O= Mass /molar mass

Number of moles of H₂O= 54 g / 18/mol

Number of moles of  H₂O= 3 mol

Now we will compare the moles of water with  oxygen .

                 H₂O             :            O₂

                  2                 :             1

                  3                  :            1/2 × 3 =1.5

              

The 3 moles of H₂O produced by 1.5 moles of oxygen.

Mass of O₂  :

Mass of O₂  :  = number of moles × molar mass

Mass of O₂  : = 1.5 mol ×32 g/mol

Mass of O₂  :  = 48 g

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\text{H}_2\text{PO}_4^{-} loses one proton to produce \text{HPO}_4^{2-} in this reaction.

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Below are the reduction half reactions for chemolithoautotrophic nitrification, where ammonia is a source of electrons and energ
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<u>Answer:</u> The molecules of oxygen gas that will be reduced to water are 42 molecules

<u>Explanation:</u>

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E^o_{(NO_2^-/NH_4)}=-0.41V\\E^o_{(O_2/H_2O)}=0.82V

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction. Here, oxygen will undergo reduction reaction will get reduced.

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Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

The half reactions follows:

<u>Oxidation half reaction:</u>  NH_4\rightarrow NO_2^-+6e^-     ( × 4)

<u>Reduction half reaction:</u>  O_2+4e^-\rightarrow 2H_2O   ( × 6)

<u>Overall reaction:</u> 4NH_4+6O_2\rightarrow 4NO_2^-+12H_2O

We are given:

Molecules of NH_4 = 28

By Stoichiometry of the reaction:

4 molecules of NH_4 reacts with 6 molecules of oxygen gas

So, 28 molecules of NH_4 will react with = \frac{6}{4}\times 28=42 molecules of oxygen gas

Hence, the molecules of oxygen gas that will be reduced to water are 42 molecules

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