Question

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 permanganate solution to reach the endpoint of a titration. What is the percent of iron (55.845 g/mol) in the salt? Report Sheet Table 1: Standardization of KMnO with Sodium Oxalate Trial 2 Trial 3 Mass of sodium oxalate ( 0.19 0.099 Molar mass of sodium oxalate (g/mol) 88.0199/mol Buret reading of Mno. Initial (mL) 5.34 m2 17.28 mL Buret reading of Mno, final (mL) 18.28 mL 30.67 mL Volume of Mnoused (ml) (12.94 m2 13.39 mL Moles of C20% (mol) 0.006 mois 0.00 S mou Moles of Mnoz (mol) Molarity of Mnoz (mol/L) Average molarity of Mno(mol/L) Standard deviation of concentration (mol/L) RSD of concentration (%RSD) - Oolg oxalatex Imol x 5 mois = 0.0056m - 88.019 -0.099 oxalatex Imol x 5 mois 0.005m. 88.019

Answer 1

The moles of MnO4- are calculated:

n MnO4- = 0.006 mol oxa * 2 mol MnO4- / 1 mol oxa = 0.012 mol

Trial 2: 0.010 mol

The molarity is calculated:

[MnO4-] = n / V = ​​0.012 / 0.01294 = 0.93 M

Trial 2: 0.75 M

Average = 0.84 M

SD = 0.13 M

% RSD = SD * 100 / average = 0.13 * 100 / 0.84 = 15.5%

It has the molarity of MnO4- = 0.84 M

MnO4- moles are calculated:

n MnO4- = M * V = 0.84 M * 0.01023 L = 0.0086 mol

Trial 2: 0.0112 mol

The moles of Fe + 2 are half of those of MnO4-, the mass is calculated:

m Fe + 2 = n * MM / 2 = 0.0086 * 55.85 / 2 = 0.240 g

Trial 2: 0.313 g

The percentage of Fe + 2 is calculated:

% Fe + 2 = 0.24 * 100 / 0.325 = 73.85%

Trial 2: 78.05%

Average = 75.95%

SD = 2.97%

% RSD = 2.97 * 100 / 75.95 = 3.91%

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