Question

A 0.495 g sample of iron containing salt required 20.22 mL of 0.0194 M permanganate solution...

A 0.495 g sample of iron containing salt required 20.22 mL of 0.0194 M permanganate solution to reach the endpointof a titration. What is the percent of iron (55.845 g/mol) in the salt?

Work shown please

0 0
Add a comment Improve this question Transcribed image text
Know the answer?
Add Answer to:
A 0.495 g sample of iron containing salt required 20.22 mL of 0.0194 M permanganate solution...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • The Fe2+ (55.845 g/mol) content of a 2.264 g steel sample dissolved in 50.00 mL of...

    The Fe2+ (55.845 g/mol) content of a 2.264 g steel sample dissolved in 50.00 mL of an acidic solution was determined by tiration with a standardized 0.120 M potassium permanganate (KMnO4, 158.034 g/mol) solution. The titration required 44.82 mL to reach the end point. What is the concentration of iron in the steel sample? Express your answer as grams of Fe per gram of steel Mn2+5 Fe34 H20 + 5 Fe2+ MnO8 H concentration g Fe/g steel A 1.969 g...

  • i need help answering the table by tonight for my test!! thank you!! Trial 1 Prelaboratory...

    i need help answering the table by tonight for my test!! thank you!! Trial 1 Prelaboratory Assignment 1. What is the primary standard used in this experiment? What is the molar mass of the primary standard? 2. Use the chemical formula for the primary standard to write the chemical reaction for the standardization of potassium permanganate. 3. What will indicate the endpoint of the titration? 4. A 0.495 g sample of iron containing salt required 20.22 mL of 0.0194 M...

  • The Fe2+ (55.845 g/mol) content of a 2.370 g steel sample dissolved in 50.00 mL was...

    The Fe2+ (55.845 g/mol) content of a 2.370 g steel sample dissolved in 50.00 mL was determined by tiration with a standardized 0.120 M potassium permanganate (KMnO4, 158.034 g/mol) solution. The titration required 39.51 mL to reach the end point. What is the concentration of iron in the steel sample? Express your answer as grams of Fe per grams of steel (g Fe2+1 g steel). Mno, +8H+ + 5Fe2+ = Mn2+ + 5Fe'+ + 4H,0 Number g Fe2+1 g steel

  • A 48.50 mL aliquot from a 0.475 L solution that contains 0.495 g of MnSO 4...

    A 48.50 mL aliquot from a 0.475 L solution that contains 0.495 g of MnSO 4 ( MW = 151.00 g/mol) required 37.2 mL of an EDTA solution to reach the end point in a titration. What mass, in milligrams, of CaCO 3 ( MW = 100.09 g/mol) will react with 1.00 mL of the EDTA solution?

  • The Fe2+ (55.84555.845 g/mol) content of a 2.028g steel sample dissolved in 50.00mL of an acidic solution was determined...

    The Fe2+ (55.84555.845 g/mol) content of a 2.028g steel sample dissolved in 50.00mL of an acidic solution was determined by tiration with a standardized 0.130 M potassium permanganate (KMnO4KMnO4, 158.034 g/mol) solution. The titration required 36.61 mL to reach the end point. What is the concentration of iron in the steel sample? Express your answer as grams of Fe per gram of steel. MnO−4+8H++5Fe2+↽−−⇀Mn2++5Fe3++ concentration: ______________ g Fe/g steel

  • Question 10 (2 points) A 13.9800-g sample of a salt was analyzed for its iron content...

    Question 10 (2 points) A 13.9800-g sample of a salt was analyzed for its iron content as follows: the entire sample was dissolved and diluted to 400.0 mL in a volumetric flask. A 80.00-ml aliquot of the solution required 25.38-mL of a 0.04119 M potassium permanganate solution to reach equivalence. A blank correction of 0.50 mL was required. Calculate the %(w/w) Fe in the sample. The net-ionic equation for the reaction is MnO4 (aq) + 8H+ (aq) + 5Fe2+(aq) -...

  • 4. A 2.86 g sample containing both Fe and V was dissolved under certain conditions and diluted to 200.00 mL. Fe2 and VO ions. The titration of this solution required 22.64 mL of0.1000 M Ce" t...

    4. A 2.86 g sample containing both Fe and V was dissolved under certain conditions and diluted to 200.00 mL. Fe2 and VO ions. The titration of this solution required 22.64 mL of0.1000 M Ce" to reach end point. A second 50.00 mL aliquot was passed through a Jones reductor to forn ions. The titration of the second solution required 42.66 mL of 0.1000 M Ce solution to reach an end point. Calculate the percentage of Fe and V in...

  • 1.) A 26.7 mL sample of a 0.495 M aqeuous hydrofluoric acid solution is titrated with...

    1.) A 26.7 mL sample of a 0.495 M aqeuous hydrofluoric acid solution is titrated with a 0.334 M aqeuous sodium hydroxide solution. what is the pH at the start of the titration, before any sodium hydroxide has been added? pH = ???? 2.) what is the pH at the equivalence point in the titration of a 18.3 mL sample of a 0.429 M aqueous hypochlorous acid solution with a 0.479 M aqueous barium hydroxide solution? pH = ???

  • 3. A 1.6152 g sample of a solution containing H2O, required 11.78 mL of a 0.6134...

    3. A 1.6152 g sample of a solution containing H2O, required 11.78 mL of a 0.6134 M KMnO4 solution to reach the equivalence point. a. How many grams of H2O2 were present in the solution? b. What is the percent H2O2 present in the solution? X 100 1.6152 93

  • Question 5 2/2 points A 3.7602 g sample of an unknown salt containing ferrous ion was...

    Question 5 2/2 points A 3.7602 g sample of an unknown salt containing ferrous ion was dissolved and diluted to 250.0 ml. Repeat 25.00 mL samples of the total solution were titrated with 0.01221 M KMNO4 solution, and the mean of three accepted corrected titration volumes was 13.36 mL. Calculate the %w/w of iron in the original sample mass. State your answer to 2 places after the decimal place. Do not enter units.

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT