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Travka [436]
3 years ago
9

Which equation represents an oxidation-reduction reaction?

Chemistry
1 answer:
bazaltina [42]3 years ago
3 0
In an oxidation-reduction reaction there is an exchange of electrons.


The exchange of electrons implies change in the oxidation states: at least one element increases its oxidation number while other reduces it.


By simple ispection you can predict that in the equation b. there is a change in oxidation states of Cl and Mn.


Now you can check it:


Equation             4H   Cl     +    Mn  O2     ->    Mn  Cl2   + 2H2 O     + Cl2    

oxidation sates   1+  1-             4+   2-              2+   1-         1+   2-         0


The oxidation state of Cl in HCl is 1-  and it changed to 0 in Cl2


The oxidation state of Mn in MnO2 is 4+ and it changed to 2+ in MnCl2 


Answer b.   
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Darya [45]

Answer:

See Explanation

Explanation:

The equation of the reaction;

KHSO4(aq) + KOH(aq) -------> K2SO4(aq) + H2O(l)

Number of moles of KHSO4 = 49.6 g/136.169 g/mol = 0.36 moles

Since the reaction is in a mole ratio of 1:1, 0.36 moles of K2SO4 is produced.

Number of moles of KOH = 25.3 g/56.1056 g/mol = 0.45 moles

Since the reaction is 1:1, 0.45 moles of K2SO4 is produced

Hence K2SO4 is the limiting reactant.

Mass of K2SO4 formed = 0.36 moles of K2SO4 * 174.26 g/mol = 62.7 g

So;

1 mole of KHSO4 reacts with 1 mole of KOH

0.36 moles of KHSO4 reacts with 0.36 * 1/1 = 0.36 moles of KOH

Amount of excess KOH = 0.45 moles - 0.36 moles = 0.09 moles

Mass of excess KOH = 0.09 moles * 56.1056 g/mol  = 5 g of excess KOH

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