
Oct 21, 2019 Solution for A 0.8040 gram sample of an iron ore is dissolved in acid. The iron is then reduced to Fe2+ and titrated with 47.22 mL of 0.02242 M KMnO4 solution.

The iron content of iron ore can be determined by titration with a standard KMnO 4 solution. The iron ore is dissolved in HC1, and all the iron is reduced to Fe 2+ ions. This solution is then titrated with KMn0 4 solution, producing Fe 3+ and Mn 2+ ions in acidic solution. If it required 38.37 mL of 0.0198 M KMnO 4 to titrate a solution made from 0.6128 g of iron ore, what is the mass percent ...

Boiling Point (BP), Iron changes its state from liquid to gas at 2861°C (5181.8°F or 3134.15K) Iron is a soft, silver-white or gray, and ductile metal. Symbol: Fe [ element] Atomic number: 26 Atomic weight: 55.845(2) Ionization energy, eV: 7.9024678. Molecular formula: Fe. Molecular weight: 55.845 g/mol. Molar volume: 7.093 cm³/mol

The mass percent of iron in an iron ore can be determined by converting Fe to Fe2+ in acid solution, and then titrating the dissolved sample with KMnO4 solution. A 2.51 g ore sample dissolved in acid required 28.62 mL of 0.0821 M KMnO4 to reach the equivalence point. Calculate the mass percent of iron in the ore.

An iron nail with a mass of 0.750 g was dissolved in dilute sulfuric acid and diluted to 100.0 mL volume. A 10.0 mL aliquot of this solution was then titrated with standard 0.200 M potassium permanganate solution, according to the following equation: MnO4-


Mass of impure Iron(II) salt = 1.322g Table of standardization of the unknown sample Volume (mL)Volume (mL)Volume (mL) 0.6119.3228.21 Initial Burette reading10.4110.4119.32 Volume of KMnO49.808.918.89

The amount of iron in a meteorite was determined by a redox titration using KMnO4 as the titrant. A 0.4408 g sample was dissolved in acid and the liberated Fe3+ quantitatively reduced to Fe2+. Titrating with 0.0220 M KMnO4 requires 41.38 ml to reach the end point. Unbalance reaction: Fe2+ + MnO4- → Fe3+ + Mn2+. Determine the % w/w Fe2O3.

(ii) The mass of iron in the sample, in grams mass Fe = 5 × M × V × 55.85 g mol−1 OR mass Fe = mol Fe2+ × 55.85 g mol−1 One point is earned for the answer in part (d)(i) multiplied by 55.85. (iii) The mass percent of iron in the compound mass % Fe = 5 55.85×××MV g × 100 OR mass % Fe = mass Fe g × 100 One point is earned for the ...

(0.0195 mol KMnO4 / 1L) X 0.02435= 0.000475 mol KMnO4 Step 3: use the stoichiometric factor to calculate the amount of iron (ii) ion 0.000475 mol KMnO4 X (5 mol Fe / 1 mol KMnO4) = 0.00237 mol Fe Step 4: calculate the mass of iron 0.00237 mol Fe X (55.85g Fe / 1 mol Fe) = 0.133g Fe Step 5: determine the mass percent 0.133g Fe / 1.026 g sample ...

For 0.02 M titrant and assuming 50 mL burette, aliquot taken for titration should contain about 0.2-0.25 g of iron (3.5-4.5 millimoles). end point detection As usual in the case of permanganate titrations, pink color of excess permanganate is strong enough so that there is

A sample of iron ore weighing 0.2792 g was dissolved in an excess of a dilute acid solution. All the iron was first converted to Fe(II) ions. The solution then required 23.30 mL of 0.0194 M KMnO 4 for oxidation to Fe(III) ions. Calculate the percent by mass of iron in the ore.

The mass and percent by mass of Fe+2 in the unknown is then calculated. Procedure Standardization of KMnO4 Solution Prepare a 1 M H2SO4 solution by slowly pouring 20 mL of concentrated (18M) H2SO4 into 340 - 350 mL of distilled water in a 500 mL Erlenmeyer flask. Swirl to insure thorough

››More information on molar mass and molecular weight. In chemistry, the formula weight is a quantity computed by multiplying the atomic weight (in atomic mass units) of each element in a chemical formula by the number of atoms of that element present in the formula, then adding all of these products together.

A 1.45 g sample of an iron ore is dissolved in acid and the iron obtained is Fe2+(aq). To titrate the solution, 21.6 mL of 0.102 M KMnO4(aq) is required. What is the percent of iron in the ore? Hint: calculate moles of iron then mass of iron then % iron. 5Fe2+ + MnO4- + 8H+ --> 5Fe2+ + Mn2+ + 4H2O a. 57.6% b. 42.4% c.

Data processing. Moles of manganate(VII) = 0.01 x 0.01 = 0.0001 mol. Therefore, from the reaction stoichiometry, moles of iron(II) = 5 x 0.0001 = 5.00 x 10-4 mol. This is in 27.90 ml, therefore in 250 ml there is (250/27.9) x 5.00 x 10-4 = 8.96 x 5 x 10-4 moles of iron = 4.48 x 10-3 mol. Therefore mass of iron in 250 ml = 56 x 4.48 x 10-3 g = 0.25 g. Hence, mass of iron per tablet = 250 mg

2. Knowing the molarity of your Fe(II) solution then allows you to calculate the number of moles of iron (as Fe 2+) in your unknown sample and the mass of Fe in g: mass(Fe) = (moles Fe) ´ (55.85 g Fe/mole Fe) m(Fe)=55.85 x (total volume) x molarity(Fe(II)) Take into account that your total volume is 100 mL and atomic weight of Fe is 55.85. 5.

Q2. The above standardized KMnO4 solution was used to titrate the oxalic acid (H2C2O4, this is not a hydrate) in rhubarb leaves. A 12.38 g sample of rhubarb leaves was extracted with water and the resulting oxalic acid solution was titrated with 36.56 mL of the KMnO4 solution. What is the percent oxalic acid in the rhubarb.

Problem #2: Potassium dichromate is used to titrate a sample containing an unknown percentage of iron. The sample is dissolved in H 3 PO 4 /H 2 SO 4 mixture to reduce all of the iron to Fe 2+ ions. The solution is then titrated with 0.01625 M K 2 Cr 2 O 7, producing Fe 3+ and Cr 3+ ions in acidic solution. The titration requires 32.26 mL of K 2 Cr 2 O 7 for 1.2765 g of the sample.

Dilute the solution to volume. Transfer four 25 mL aliquots of the solution to four 200 mL Erlenmeyer flasks. Warm the solution gently and perform the pre‐reduction of the iron as described above. Once the samples have been pre‐reduced, you may proceed with the titration of the iron ore samples with Ce4+ as described above.

Redox Titration Lab ABSTRACT: In this lab, 0.010 M purple-colored potassium permanganate solution was standardized by redox titration with iron (II) ammonium sulfate hexahydrate (FAS). The average mass of the three flasks of FAS was 0.483 grams. Once the concentration of the standard solution of KMnO4 (aq) was determined, it was used to determine the concentration of Fe2+ in iron pills.

››More information on molar mass and molecular weight. In chemistry, the formula weight is a quantity computed by multiplying the atomic weight (in atomic mass units) of each element in a chemical formula by the number of atoms of that element present in the formula, then adding all of these products together.

The iron ore is dissolved in HCl and all the iron is reduced to Fe^2+ ions. The solution is then titrated with KMnO4 solution, producing Fe^3+ and Mn^2+ ions in acidic solution. If it required 41.95 mL of 0.0205M KMnO4 to titrate a solution made from 0.6128 g of iron ore, what is the percent (by mass) of iron in the iron ore?

Q2. The above standardized KMnO4 solution was used to titrate the oxalic acid (H2C2O4, this is not a hydrate) in rhubarb leaves. A 12.38 g sample of rhubarb leaves was extracted with water and the resulting oxalic acid solution was titrated with 36.56 mL of the KMnO4 solution. What is the percent oxalic acid in the rhubarb.

Answer to I cannot figure out how to solve these questions. I have reviewed my textbook, watched video after video I keep gettin...

By determining the exact mass of the FAS samples taken, and from the volume of KMnO 4 solution required to titrate those samples, the exact molarity of the KMnO 4 solutions could be calculate.. In the second week of the experiment, you titrated samples of an "unknown" iron(II) salt with the same KMnO 4 solution, in order to determine what % by mass of Fe it contained.

Problem #2: Potassium dichromate is used to titrate a sample containing an unknown percentage of iron. The sample is dissolved in H 3 PO 4 /H 2 SO 4 mixture to reduce all of the iron to Fe 2+ ions. The solution is then titrated with 0.01625 M K 2 Cr 2 O 7, producing Fe 3+ and Cr 3+ ions in acidic solution. The titration requires 32.26 mL of K 2 Cr 2 O 7 for 1.2765 g of the sample.

Dilute the solution to volume. Transfer four 25 mL aliquots of the solution to four 200 mL Erlenmeyer flasks. Warm the solution gently and perform the pre‐reduction of the iron as described above. Once the samples have been pre‐reduced, you may proceed with the titration of the iron ore samples with Ce4+ as described above.

2. Knowing the molarity of your Fe(II) solution then allows you to calculate the number of moles of iron (as Fe 2+) in your unknown sample and the mass of Fe in g: mass(Fe) = (moles Fe) ´ (55.85 g Fe/mole Fe) m(Fe)=55.85 x (total volume) x molarity(Fe(II)) Take into account that your total volume is 100 mL and atomic weight of Fe is 55.85. 5.

Calculate the concentration of the iron(II) solution, and from this calculate the mass of iron in an iron tablet. Specimen results Mass of iron tablets = 1.81 g Rough titre = 17.0 cm 3 Second titre = 16.7 cm 3 Third titre = 16.7 cm 3 Average of accurate titres = 16.7 cm 3 Volume of iron(II) solution used in each titration = 25.0 cm 3

Ex. 13-7. A 0.8040-g sample of an iron ore is dissolved in acid. The iron is then reduced to Fe2+ and titrated with 47.22 mL of 0.02242 M KMnO 4 solution. Calculate the results of this analysis in term of (a) % Fe (55.847 g/mol); and (b) % Fe3O4 (231.54 g/mol). MnO4-+ 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4H 2O (a). stoichiometric ratio = 4 2 1 mmol ...

A 1.45 g sample of an iron ore is dissolved in acid and the iron obtained is Fe2+(aq). To titrate the solution, 21.6 mL of 0.102 M KMnO4(aq) is required.

• The determination of iron in an ore: Section 36E-4. Use a first sample size of about 1 g if your unknown is < 30% iron and 0.6 g if > 30% iron. The sample size should then be adjusted so that the titration volume is about 40 mL. Preparation of 0.02 M potassium permanganate.

NB let say i got a volume of 35 ml after the titration and the weight of the flask was 180 g. how can i get the mass of iron .. View How do I measure both ferrous and ferric iron in water in a ...

1. Volume of HCl used (10 mL) mL mL mL 2. Initial buret reading (NaOH) mL mL mL 3. Final buret reading (NaOH) mL mL mL Back-titration of limestone Trial 1 Trial 2 7. Mass of limestone g g 8. Volume of HCl added (50 mL) mL mL 9. Initial buret reading (NaOH) mL mL 10. Final buret reading (NaOH) mL mL