Class 11 Exam > Class 11 Questions > The number of moles of KMnO4 reduced by one m...
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The number of moles of KMnO4 reduced by one mole of KI in alkaline medium is: [2 00 5]
- a)one
- b)two
- c)five
- d)one fifth
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
The number of moles of KMnO4 reduced by one mole of KI in alkaline med...
Change in oxidation number of Mn in basic medium is 1. Hence mole of KI is equal to mole of KMnO4.
Most Upvoted Answer
The number of moles of KMnO4 reduced by one mole of KI in alkaline med...
Given reaction: KMnO4+KI→MnO2+KIO3
In a balanced chemical reaction number of all atoms at right side should be equal to the left side of the reaction.
Balancing a chemical reaction as:
Step 1: Assign oxidation numbers to each of the atoms in the equation and write the numbers above the atom as:
KMn+7O4+KI−1→Mn+4O2+KI+5O3
Step 2: Identify the atoms that are oxidized and those that are reduced as:
Reduction: KMn+7O4→Mn+4O2
Oxidation: KI−1→KI+5O3
Step 3: oxidation-number change is:
Reduction: KMn+7O4→Mn+4O2:gain of total 3 electrons
Oxidation: KI−1→KI+5O3 Loss of total 6 electrons
Step 4: Balance the total change in oxidation number as:
Reduction: KMn+7O4→Mn+4O2×2: gain of total 6 electrons
Oxidation: KI−1→KI+5O3×1 Loss of total 6 electrons
∴ Reduction: 2KMn+7O4→2Mn+4O2
Oxidation: KI−1→KI+5O3
Step 5: Balance O atoms in reduction reaction by adding H2O and then balance H by H+ as:
Reduction: 2KMn+7O4+8H+→2Mn+4O2+4H2O+2K+
Oxidation:
Step 6: For base catalysed reaction add OH− to both side to neutralize H+ as:
2KMn+7O4+8H++8OH−→2Mn+4O2+8OH−+2K++4H2O
Oxidation: KI−1+6OH−+3H2O→KI+5O3+6H++6OH−
or Reduction: 2KMn+7O4+4H2O→2Mn+4O2+8OH−+2K+
Oxidation: KI−1+6OH−→KI+5O3+3H2O
Thus overall reaction is:
2KMnO4+4H2O+
In a balanced chemical reaction number of all atoms at right side should be equal to the left side of the reaction.
Balancing a chemical reaction as:
Step 1: Assign oxidation numbers to each of the atoms in the equation and write the numbers above the atom as:
KMn+7O4+KI−1→Mn+4O2+KI+5O3
Step 2: Identify the atoms that are oxidized and those that are reduced as:
Reduction: KMn+7O4→Mn+4O2
Oxidation: KI−1→KI+5O3
Step 3: oxidation-number change is:
Reduction: KMn+7O4→Mn+4O2:gain of total 3 electrons
Oxidation: KI−1→KI+5O3 Loss of total 6 electrons
Step 4: Balance the total change in oxidation number as:
Reduction: KMn+7O4→Mn+4O2×2: gain of total 6 electrons
Oxidation: KI−1→KI+5O3×1 Loss of total 6 electrons
∴ Reduction: 2KMn+7O4→2Mn+4O2
Oxidation: KI−1→KI+5O3
Step 5: Balance O atoms in reduction reaction by adding H2O and then balance H by H+ as:
Reduction: 2KMn+7O4+8H+→2Mn+4O2+4H2O+2K+
Oxidation:
Step 6: For base catalysed reaction add OH− to both side to neutralize H+ as:
2KMn+7O4+8H++8OH−→2Mn+4O2+8OH−+2K++4H2O
Oxidation: KI−1+6OH−+3H2O→KI+5O3+6H++6OH−
or Reduction: 2KMn+7O4+4H2O→2Mn+4O2+8OH−+2K+
Oxidation: KI−1+6OH−→KI+5O3+3H2O
Thus overall reaction is:
2KMnO4+4H2O+
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Community Answer
The number of moles of KMnO4 reduced by one mole of KI in alkaline med...
To determine the number of moles of KMnO4 reduced by one mole of KI in alkaline medium, we need to understand the balanced chemical equation for the reaction and the stoichiometry involved.
The balanced chemical equation for the reaction between KMnO4 and KI in alkaline medium is as follows:
2 KMnO4 + 10 KOH + 5 KI -> 2 K2MnO4 + 6 K2CO3 + 3 I2 + 5 H2O
From the balanced equation, we can see that 2 moles of KMnO4 react with 5 moles of KI. Therefore, the stoichiometric ratio between KMnO4 and KI is 2:5.
Now, we need to determine the number of moles of KMnO4 reduced by one mole of KI. Since the stoichiometric ratio is 2:5, we can set up a ratio:
2 moles KMnO4 / 5 moles KI = x moles KMnO4 / 1 mole KI
Simplifying this ratio, we find:
x = (2/5) * 1 = 2/5
Therefore, the number of moles of KMnO4 reduced by one mole of KI in alkaline medium is 2/5. This can also be expressed as 0.4 moles of KMnO4.
Therefore, the correct answer is option A) two.
The balanced chemical equation for the reaction between KMnO4 and KI in alkaline medium is as follows:
2 KMnO4 + 10 KOH + 5 KI -> 2 K2MnO4 + 6 K2CO3 + 3 I2 + 5 H2O
From the balanced equation, we can see that 2 moles of KMnO4 react with 5 moles of KI. Therefore, the stoichiometric ratio between KMnO4 and KI is 2:5.
Now, we need to determine the number of moles of KMnO4 reduced by one mole of KI. Since the stoichiometric ratio is 2:5, we can set up a ratio:
2 moles KMnO4 / 5 moles KI = x moles KMnO4 / 1 mole KI
Simplifying this ratio, we find:
x = (2/5) * 1 = 2/5
Therefore, the number of moles of KMnO4 reduced by one mole of KI in alkaline medium is 2/5. This can also be expressed as 0.4 moles of KMnO4.
Therefore, the correct answer is option A) two.
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