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100ml of 0.1n feso4 solution will be compelety oxidised by x gms of k2cr207 in acidic medium .the value of x is?
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100ml of 0.1n feso4 solution will be compelety oxidised by x gms of k2...
Introduction
To determine the amount of K2Cr2O7 required to completely oxidize 100 ml of 0.1 N FeSO4 solution in an acidic medium, we follow these steps:
Step 1: Calculate the moles of FeSO4
- Normality (N) of FeSO4 = 0.1 N
- Volume (V) = 100 ml = 0.1 L
- Moles of FeSO4 = Normality × Volume = 0.1 × 0.1 = 0.01 equivalents
Step 2: Determine the equivalents of Fe2+
- Each mole of FeSO4 provides one equivalent of Fe2+.
- Thus, equivalents of Fe2+ = 0.01 eq.
Step 3: Stoichiometry of the reaction
- The balanced redox reaction in acidic medium is:
6 Fe2+ + Cr2O7^2- + 14 H+ → 6 Fe3+ + 2 Cr3+ + 7 H2O
- From the equation, 1 equivalent of Cr2O7^2- oxidizes 6 equivalents of Fe2+.
Step 4: Calculate equivalents of K2Cr2O7 needed
- Since 6 equivalents of Fe2+ require 1 equivalent of Cr2O7^2-, we can set up the proportion:
- For 0.01 eq of Fe2+, the equivalents of Cr2O7^2- needed = 0.01 / 6 = 0.00167 eq.
Step 5: Calculate grams of K2Cr2O7
- Molar mass of K2Cr2O7 = 294 g/mol
- Since 1 mol of K2Cr2O7 = 6 equivalents, the equivalent weight = 294 / 6 = 49 g/eq.
- Thus, mass of K2Cr2O7 needed = 0.00167 eq × 49 g/eq = 0.08183 g.
Conclusion
The value of x, the mass of K2Cr2O7 required, is approximately 0.0818 g.
To determine the amount of K2Cr2O7 required to completely oxidize 100 ml of 0.1 N FeSO4 solution in an acidic medium, we follow these steps:
Step 1: Calculate the moles of FeSO4
- Normality (N) of FeSO4 = 0.1 N
- Volume (V) = 100 ml = 0.1 L
- Moles of FeSO4 = Normality × Volume = 0.1 × 0.1 = 0.01 equivalents
Step 2: Determine the equivalents of Fe2+
- Each mole of FeSO4 provides one equivalent of Fe2+.
- Thus, equivalents of Fe2+ = 0.01 eq.
Step 3: Stoichiometry of the reaction
- The balanced redox reaction in acidic medium is:
6 Fe2+ + Cr2O7^2- + 14 H+ → 6 Fe3+ + 2 Cr3+ + 7 H2O
- From the equation, 1 equivalent of Cr2O7^2- oxidizes 6 equivalents of Fe2+.
Step 4: Calculate equivalents of K2Cr2O7 needed
- Since 6 equivalents of Fe2+ require 1 equivalent of Cr2O7^2-, we can set up the proportion:
- For 0.01 eq of Fe2+, the equivalents of Cr2O7^2- needed = 0.01 / 6 = 0.00167 eq.
Step 5: Calculate grams of K2Cr2O7
- Molar mass of K2Cr2O7 = 294 g/mol
- Since 1 mol of K2Cr2O7 = 6 equivalents, the equivalent weight = 294 / 6 = 49 g/eq.
- Thus, mass of K2Cr2O7 needed = 0.00167 eq × 49 g/eq = 0.08183 g.
Conclusion
The value of x, the mass of K2Cr2O7 required, is approximately 0.0818 g.
Community Answer
100ml of 0.1n feso4 solution will be compelety oxidised by x gms of k2...
Is it 0.62 gram
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100ml of 0.1n feso4 solution will be compelety oxidised by x gms of k2cr207 in acidic medium .the value of x is?
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100ml of 0.1n feso4 solution will be compelety oxidised by x gms of k2cr207 in acidic medium .the value of x is? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about 100ml of 0.1n feso4 solution will be compelety oxidised by x gms of k2cr207 in acidic medium .the value of x is? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 100ml of 0.1n feso4 solution will be compelety oxidised by x gms of k2cr207 in acidic medium .the value of x is?.
100ml of 0.1n feso4 solution will be compelety oxidised by x gms of k2cr207 in acidic medium .the value of x is? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about 100ml of 0.1n feso4 solution will be compelety oxidised by x gms of k2cr207 in acidic medium .the value of x is? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 100ml of 0.1n feso4 solution will be compelety oxidised by x gms of k2cr207 in acidic medium .the value of x is?.
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