A sample of an unknown containing Fe2+ in the dissolved sample requires 23.15 mL of a 0.0020 M KMnO4 solution to reach the end point of the titration. 1)Calculate the moles of KMnO4 reacted. 2) Based on the moles you calculated , calculate how many grams of Fe2+ are in the unknown solution.
A sample of an unknown containing Fe2+ in the dissolved sample requires 23.15 mL of a...
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...
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
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. (4p) In an acidic aqueous solution, Fe2+ ions are oxidized to Fe3+ ions by MnO4": 5Fe2+(aq) + MnO4 (aq) + 8H(aq) → 5Fe3+ (aq) + Mn2(aq) + 4H2O(1) In part A of the experiment, suppose that 1.067 g of Fe(NH4)2(SO4)2.6H2O(s) are placed in a 250 ml Erlenmeyer flask to which 20 mL of water and 8 mL of 3 M H2SO4(aq) are added. The solution was titrated to the end point by adding 26.89 mL of KMnO4(aq) from the...
suppose you have an unknown sample containing Fe^2+, and you titrate it with KMnO4 standard solution. Calculate the percent of Fe in your sample. equation: MnO4- + 5Fe^2+ + 8H^+ --> Mn^2+ + 5Fe^3+ + 4H2O Suppose you have an unknown sample containing Feat, and you titrate it with a KMnO4 standard solution. Calculate the percentage of Fe in your sample. Equation: MnO(+5Fe2+ + 8H - Mn2+ + 5Fe3+ + 4H20 Data 0.08124M 2.49g Molarity of the KMnO4 standard solution:...
4. A 0.5255 g sample containing an unknown mass of SnCl2 is dissolved in 50.00 mL of water and titrated with 0.02050 M KMnO4 under mildly acidic conditions. The unbalanced titration reaction is shown below. MnO4 (aq) + Sn²(aq) MnO2(8) + Sn** (aq) a. Use the half-reaction method to balance the titration reaction. (5 pts) b. Given that 32.25 mL of KMnO, was required to reach the endpoint, calculate the percent SnCl2 in the sample. (5 pts)
3. 0.7253 g sample containing an unknown weak acid HA was dissolved in 50 mL water and titrated against 0.1555 M NaOH, requiring 48.11 mL to of NaOH to reach the end-point. During the titration reaction, the pH of the solution is 3.77 when half of the HA is neutralized and the equivalence-point pH is 8.33. (a) State two ways to standardize the NaOH used in the titration. (b) Suggest and explain an indicator that can be used in the...
Problem 10.30 A 21.5 mL sample of a KOH solution of unknown concentration requires 11.1 mL of 0.250 M H2So4 solution to reach the end point in a titration.
The student dissolved 2.050 grams of acid in 50 mL of distilled water. He then titrated the unknown acid with a standardized NaOH solution. Unfortunately, the student over shot the end point and had to back titrate the solution with 2.17 mL of 0.98861 M HCl. Once the final end point was determined, the student had added a total of 29.15 mL of a 1.0324 M NaOH solution. A) How many moles of OH- were used to reach the end...
The student dissolved 2.050 grams of acid in 50 mL of distilled water. He then titrated the unknown acid with a standardized NaOH solution. Unfortunately, the student over shot the end point and had to back titrate the solution with 2.17 mL of 0.98861 M HCl. Once the final end point was determined, the student had added a total of 29.15 mL of a 1.0324 M NaOH solution. A) How many moles of OH- were used to reach the end...