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100 ml of unknown reducing agent is reacted with 500 mL, 0.1 M excess I2 solution. If remaining I2 required 0.1 M, 200 mL Na2S2O3 solution for complete neutralization. Calculate normality of unknown reducing agent.
Correct answer is '0.8'. Can you explain this answer?
Most Upvoted Answer
100 ml of unknown reducing agent is reacted with 500 mL, 0.1 M excess ...
Given information:
- Volume of unknown reducing agent = 100 mL
- Volume of excess I2 solution = 500 mL
- Concentration of excess I2 solution = 0.1 M
- Volume of Na2S2O3 solution required = 200 mL
- Concentration of Na2S2O3 solution = 0.1 M
- Normality of unknown reducing agent = ?
To find the normality of the unknown reducing agent, we need to use the equation:
Normality of reducing agent x Volume of reducing agent = Normality of oxidizing agent x Volume of oxidizing agent
where reducing agent is the unknown substance, and oxidizing agent is I2.
Step 1: Determine the moles of I2 present in the excess solution.
Molarity of I2 = 0.1 M
Volume of I2 solution = 500 mL = 0.5 L
Moles of I2 = Molarity x Volume = 0.1 x 0.5 = 0.05 moles
Step 2: Determine the moles of Na2S2O3 required to react with the remaining I2.
Molarity of Na2S2O3 = 0.1 M
Volume of Na2S2O3 solution = 200 mL = 0.2 L
Moles of Na2S2O3 = Molarity x Volume = 0.1 x 0.2 = 0.02 moles
Step 3: Determine the moles of I2 that reacted with Na2S2O3.
From the balanced chemical equation:
I2 + 2Na2S2O3 → 2NaI + Na2S4O6
1 mole of I2 reacts with 2 moles of Na2S2O3
Therefore, moles of I2 reacted = 0.02 x 0.5/2 = 0.005 moles
Step 4: Determine the moles of I2 that reacted with the reducing agent.
Moles of I2 initially present = 0.05 moles
Moles of I2 that reacted with Na2S2O3 = 0.005 moles
Moles of I2 that reacted with reducing agent = 0.05 - 0.005 = 0.045 moles
Step 5: Calculate the normality of the unknown reducing agent.
Normality of reducing agent x Volume of reducing agent = Normality of oxidizing agent x Volume of oxidizing agent
Normality of reducing agent x 0.1 L = 0.05 x 0.5 L
Normality of reducing agent = (0.05 x 0.5)/0.1 = 0.25 N
But the reducing agent reacted with only 0.045 moles of I2, so we need to adjust the normality accordingly.
Adjusted normality of reducing agent = (0.045/0.1) x 0.25 = 0.1125 N
Finally, we need to convert normality to molarity because the unknown reducing agent is not a strong acid or base.
Molarity = Normality x Equivalent weight
The equivalent weight of the reducing agent is half its molecular weight because it donates 1 electron per molecule.
Molecular weight of reducing agent = 2 x Equivalent weight
Therefore
- Volume of unknown reducing agent = 100 mL
- Volume of excess I2 solution = 500 mL
- Concentration of excess I2 solution = 0.1 M
- Volume of Na2S2O3 solution required = 200 mL
- Concentration of Na2S2O3 solution = 0.1 M
- Normality of unknown reducing agent = ?
To find the normality of the unknown reducing agent, we need to use the equation:
Normality of reducing agent x Volume of reducing agent = Normality of oxidizing agent x Volume of oxidizing agent
where reducing agent is the unknown substance, and oxidizing agent is I2.
Step 1: Determine the moles of I2 present in the excess solution.
Molarity of I2 = 0.1 M
Volume of I2 solution = 500 mL = 0.5 L
Moles of I2 = Molarity x Volume = 0.1 x 0.5 = 0.05 moles
Step 2: Determine the moles of Na2S2O3 required to react with the remaining I2.
Molarity of Na2S2O3 = 0.1 M
Volume of Na2S2O3 solution = 200 mL = 0.2 L
Moles of Na2S2O3 = Molarity x Volume = 0.1 x 0.2 = 0.02 moles
Step 3: Determine the moles of I2 that reacted with Na2S2O3.
From the balanced chemical equation:
I2 + 2Na2S2O3 → 2NaI + Na2S4O6
1 mole of I2 reacts with 2 moles of Na2S2O3
Therefore, moles of I2 reacted = 0.02 x 0.5/2 = 0.005 moles
Step 4: Determine the moles of I2 that reacted with the reducing agent.
Moles of I2 initially present = 0.05 moles
Moles of I2 that reacted with Na2S2O3 = 0.005 moles
Moles of I2 that reacted with reducing agent = 0.05 - 0.005 = 0.045 moles
Step 5: Calculate the normality of the unknown reducing agent.
Normality of reducing agent x Volume of reducing agent = Normality of oxidizing agent x Volume of oxidizing agent
Normality of reducing agent x 0.1 L = 0.05 x 0.5 L
Normality of reducing agent = (0.05 x 0.5)/0.1 = 0.25 N
But the reducing agent reacted with only 0.045 moles of I2, so we need to adjust the normality accordingly.
Adjusted normality of reducing agent = (0.045/0.1) x 0.25 = 0.1125 N
Finally, we need to convert normality to molarity because the unknown reducing agent is not a strong acid or base.
Molarity = Normality x Equivalent weight
The equivalent weight of the reducing agent is half its molecular weight because it donates 1 electron per molecule.
Molecular weight of reducing agent = 2 x Equivalent weight
Therefore
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Community Answer
100 ml of unknown reducing agent is reacted with 500 mL, 0.1 M excess ...
Solution:
Given:
Volume of reducing agent = 100 mL = 0.1 L
Volume of I2 solution = 500 mL = 0.5 L
Concentration of I2 solution = 0.1 M
Volume of Na2S2O3 solution = 200 mL = 0.2 L
Concentration of Na2S2O3 solution = 0.1 M
Let us first balance the chemical equation for the reaction between the reducing agent and I2:
Reducing agent + I2 → Product
Since the reducing agent is not known, we cannot write its formula. However, we can write the formula for the oxidized form of the reducing agent:
Oxidized form of reducing agent + I2 → Product
Let the normality of the oxidized form of the reducing agent be n.
We know that the volume of I2 solution is in excess, which means that all the I2 will react with the reducing agent. Therefore, the number of moles of I2 in the solution is:
n(I2) = CV = 0.1 × 0.5 = 0.05 moles
Since the reducing agent completely reacts with I2, the number of moles of the oxidized form of the reducing agent is also 0.05.
The reaction between I2 and Na2S2O3 is:
I2 + 2Na2S2O3 → 2NaI + Na2S4O6
We know the concentration and volume of Na2S2O3 solution used, so we can calculate the number of moles of Na2S2O3 used:
n(Na2S2O3) = CV = 0.1 × 0.2 = 0.02 moles
Since each mole of Na2S2O3 reacts with one mole of I2, the number of moles of I2 used is also 0.02.
Therefore, the number of moles of I2 that reacted with the reducing agent is:
n(I2) = 0.05 - 0.02 = 0.03 moles
Let the formula of the oxidized form of the reducing agent be A. Then the balanced chemical equation for the reaction between A and I2 is:
A + I2 → AI
Since each mole of A reacts with one mole of I2, the number of moles of A is also 0.03.
We know that the normality of A is n, and the volume of A used is 0.1 L. Therefore, the number of equivalents of A used is:
Equivalents of A = n × 0.1
Since each mole of A contains one equivalent of reducing power, the number of moles of A used is also n × 0.1. Therefore, we have:
n × 0.1 = 0.03
n = 0.3
Therefore, the normality of the reducing agent is:
Normality = n/2 = 0.3/2 = 0.15
However, this is the normality of the oxidized form of the reducing agent (A). To find the normality of the reducing agent itself, we need to know how many moles of A are equivalent to one mole of the reducing agent.
Given:
Volume of reducing agent = 100 mL = 0.1 L
Volume of I2 solution = 500 mL = 0.5 L
Concentration of I2 solution = 0.1 M
Volume of Na2S2O3 solution = 200 mL = 0.2 L
Concentration of Na2S2O3 solution = 0.1 M
Let us first balance the chemical equation for the reaction between the reducing agent and I2:
Reducing agent + I2 → Product
Since the reducing agent is not known, we cannot write its formula. However, we can write the formula for the oxidized form of the reducing agent:
Oxidized form of reducing agent + I2 → Product
Let the normality of the oxidized form of the reducing agent be n.
We know that the volume of I2 solution is in excess, which means that all the I2 will react with the reducing agent. Therefore, the number of moles of I2 in the solution is:
n(I2) = CV = 0.1 × 0.5 = 0.05 moles
Since the reducing agent completely reacts with I2, the number of moles of the oxidized form of the reducing agent is also 0.05.
The reaction between I2 and Na2S2O3 is:
I2 + 2Na2S2O3 → 2NaI + Na2S4O6
We know the concentration and volume of Na2S2O3 solution used, so we can calculate the number of moles of Na2S2O3 used:
n(Na2S2O3) = CV = 0.1 × 0.2 = 0.02 moles
Since each mole of Na2S2O3 reacts with one mole of I2, the number of moles of I2 used is also 0.02.
Therefore, the number of moles of I2 that reacted with the reducing agent is:
n(I2) = 0.05 - 0.02 = 0.03 moles
Let the formula of the oxidized form of the reducing agent be A. Then the balanced chemical equation for the reaction between A and I2 is:
A + I2 → AI
Since each mole of A reacts with one mole of I2, the number of moles of A is also 0.03.
We know that the normality of A is n, and the volume of A used is 0.1 L. Therefore, the number of equivalents of A used is:
Equivalents of A = n × 0.1
Since each mole of A contains one equivalent of reducing power, the number of moles of A used is also n × 0.1. Therefore, we have:
n × 0.1 = 0.03
n = 0.3
Therefore, the normality of the reducing agent is:
Normality = n/2 = 0.3/2 = 0.15
However, this is the normality of the oxidized form of the reducing agent (A). To find the normality of the reducing agent itself, we need to know how many moles of A are equivalent to one mole of the reducing agent.
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100 ml of unknown reducing agent is reacted with 500 mL, 0.1 M excess I2 solution. If remaining I2 required 0.1 M, 200 mL Na2S2O3 solution for complete neutralization. Calculate normality of unknown reducing agent.Correct answer is '0.8'. Can you explain this answer? for Chemistry 2025 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about 100 ml of unknown reducing agent is reacted with 500 mL, 0.1 M excess I2 solution. If remaining I2 required 0.1 M, 200 mL Na2S2O3 solution for complete neutralization. Calculate normality of unknown reducing agent.Correct answer is '0.8'. Can you explain this answer? covers all topics & solutions for Chemistry 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 100 ml of unknown reducing agent is reacted with 500 mL, 0.1 M excess I2 solution. If remaining I2 required 0.1 M, 200 mL Na2S2O3 solution for complete neutralization. Calculate normality of unknown reducing agent.Correct answer is '0.8'. Can you explain this answer?.
100 ml of unknown reducing agent is reacted with 500 mL, 0.1 M excess I2 solution. If remaining I2 required 0.1 M, 200 mL Na2S2O3 solution for complete neutralization. Calculate normality of unknown reducing agent.Correct answer is '0.8'. Can you explain this answer? for Chemistry 2025 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about 100 ml of unknown reducing agent is reacted with 500 mL, 0.1 M excess I2 solution. If remaining I2 required 0.1 M, 200 mL Na2S2O3 solution for complete neutralization. Calculate normality of unknown reducing agent.Correct answer is '0.8'. Can you explain this answer? covers all topics & solutions for Chemistry 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 100 ml of unknown reducing agent is reacted with 500 mL, 0.1 M excess I2 solution. If remaining I2 required 0.1 M, 200 mL Na2S2O3 solution for complete neutralization. Calculate normality of unknown reducing agent.Correct answer is '0.8'. Can you explain this answer?.
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