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SashulF [63]
2 years ago
7

The mineral rhodochrosite [manganese(II) carbonate, MnCO3] is a commercially important source of manganese. Write a half-reactio

n for the oxidation of the manganese in MnCO3 to MnO2 in neutral groundwater where the principal carbonate species is HCO3–. Add H2O, H+, and electrons as needed to balance the half-reaction.
Chemistry
1 answer:
aleksklad [387]2 years ago
5 0

Answer:

MnCO3 + 2H2O ⇄ MnO2  + HCO3-  + 2e- +3H+

Explanation:

<u>The</u> unbalanced equation

MnCO3 ⇄ MnO2  + HCO3-

In MnCO3, the oxidation number of Mn is +2

In Mno2, the oxidation number of Mn is +4

The change from +2 to +4 requires an addition of  2 electrons (to the right side).

MnCO3 ⇄ MnO2  + HCO3-  + 2e-

The total charge now is -3 on the right side. To balance this we add 3 hydrogen atoms on the right side.

MnCO3 ⇄ MnO2  + HCO3-  + 2e- +3H+

On the right side we have 4 hydrogen atoms in total. On the left side we have 0 hydrogen atoms. So to balance, we have to add 2H2O on the left side

MnCO3 + 2H2O ⇄ MnO2  + HCO3-  + 2e- +3H+

Now the reaction is balanced.

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rank the four gases (air, exhaled air, gas produced from the decomposition of H2O2, gas from decomposition of NaHCO3, in order o
SVEN [57.7K]

Answer: H₂O₂ (94%) > Air (23%) > Exhaled air (13%) > NaHCO₃ (0%)


Initial important note:


Although NaHCO₃ contents oxygen atoms, and you can calculate its compositoin, the resulting gas does not containg pure oxygen gas (O₂). For the comparisson it is not useful to calculate the content of oxygent atoms, but the concentration of O₂ gas. As such, the gas from NaHCO₃ contains 0% of pure O₂, that is why it is ranked last.


1) Air:


Source: internet


Approximate 23%. It is variable, because air is not a pure substance but a mixture of gases, whose compositon is not unique.


2) Exhaled air:


Source: internet.


Approximate 13%. The compositon of the air changes in our lungs, due to the respiration process: we inhale fresh air with around 23% of oxygen, part of this oxygen pass to the cells (lungs - blood - heart - cells) and then it is exhaled with a lower content of air and a greater content of CO₂


3) Air from the decomposition of H₂O₂.


In this case we can do a chemical calculation, since we can state the chemical equation of the reaction:


i) Chemical Equation:


H₂O₂ (g) → H₂ (g) + O₂ (g)


ii) mole ratio of the products 1 mol H₂ : 1 mol O₂


iii) convert moles into mass (grams)


1 mol H₂ × 2 × 1.008 g/mol = 2.016 g


1 mol O₂ × 2 × 15.999 g/mol = 31.998 g


Composition, % = [31.998 g / (2.016 g + 31.998 g) ] × 100 ≈ 94%



4) Air from the decomposition of NaHCO₃:


i) chemical equation:


2 NaHCO₃(s) → Na₂CO₃(s) + CO₂(g) + H₂O(g)


ii) mole ratio: take into account only the gases in the products:


1 mol CO₂ (g) : 1 mol H₂O


iii) mass in grams


CO₂: molar mass ia approximately 44.01 g/mol


H₂O: molar mass is approximately 18.02 g/mol


iii) Those gases although have oxygen atoms, do not hae free oxygen gas, which is what we are compariing. That means, that from the decomposition of NaHCO₃ you get 0% oxygen gas.


5) The result is:


H₂O₂ (94%) > Air (23%) > Exhaled air (13%) > NaHCO₃ (0%)

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3 years ago
Read 2 more answers
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