A 0.446-gram sample of iron ore is dissolved in HCl and the iron (Fe solid) is reduced to Fe2+. The solution is then titrated by 38.6mL of 0.105 N KMn)4. Calculate the percentage of Fe in the iron ore sample.
To calculate the percentage of Fe in the iron ore sample, we need to determine the amount of Fe that reacted with KMnO4 during titration.
Step 1: Calculate the number of moles of KMnO4 used in the titration.
Given:
Volume of KMnO4 solution (V) = 38.6 mL = 0.0386 L
Concentration of KMnO4 solution (C) = 0.105 N
Moles of KMnO4 (n) = C × V
= 0.105 N × 0.0386 L
Step 2: Determine the molar ratio between Fe2+ and KMnO4.
The balanced chemical equation for the reaction between Fe2+ and KMnO4 is:
5Fe^2+ + MnO4^- + 8H^+ -> 5Fe^3+ + Mn^2+ + 4H2O
The molar ratio of Fe2+ to KMnO4 is 5:1. This means that 1 mole of KMnO4 reacts with 5 moles of Fe2+.
Step 3: Calculate the number of moles of Fe2+ that reacted.
Since 1 mole of KMnO4 reacts with 5 moles of Fe2+, the number of moles of Fe2+ is given by:
Moles of Fe2+ (n) = (n KMnO4) × (5 moles Fe2+ / 1 mole KMnO4)
Step 4: Calculate the molecular weight of Fe.
The molecular weight of Fe is 55.845 g/mol.
Step 5: Calculate the mass of Fe in the iron ore sample.
The mass of Fe (m) is given by:
Mass of Fe (m) = Moles of Fe2+ (n) × Molecular weight of Fe
Step 6: Calculate the percentage of Fe in the iron ore sample.
The percentage of Fe is given by:
Percentage of Fe = (Mass of Fe / Mass of sample) × 100
Now, let's calculate the values.
Given:
Mass of the sample = 0.446 g
Step 1:
n = 0.105 N × 0.0386 L
Step 2:
Molar ratio = 5 moles Fe2+ / 1 mole KMnO4
Step 3:
n Fe2+ = (n KMnO4) × (5 moles Fe2+ / 1 mole KMnO4)
Step 4:
Molecular weight of Fe = 55.845 g/mol
Step 5:
m = n Fe2+ × Molecular weight of Fe
Step 6:
Percentage of Fe = (m / Mass of sample) × 100
Let's calculate the values step by step.
Step 1:
n = 0.105 N × 0.0386 L
n = 0.004053 mol KMnO4
Step 3:
n Fe2+ = (0.004053 mol KMnO4) × (5 moles Fe2+ / 1 mole KMnO4)
n Fe2+ = 0.02027 mol Fe2+
Step 5:
m = 0.02027 mol Fe2+ × 55.845 g/mol
m = 1.131 g Fe
Step 6:
Percentage of Fe = (1.131 g Fe / 0.446 g sample) × 100
Percentage of Fe = 254.02%
Therefore, the percentage of Fe in the iron ore sample is approximately 254.02%.
To calculate the percentage of Fe in the iron ore sample, we need to determine the number of moles of Fe that reacted with KMnO4 during the titration.
First, we need to find the number of moles of KMnO4 that reacted. To do this, we'll use the formula:
number of moles = concentration × volume
Given that the concentration of KMnO4 is 0.105 N (N stands for normality) and the volume of KMnO4 used is 38.6 mL (or 0.0386 L), we can calculate the number of moles of KMnO4:
moles of KMnO4 = 0.105 N × 0.0386 L = 0.004053 moles
The balanced chemical equation between KMnO4 and Fe2+ is:
5Fe2+ + MnO4- + 8H+ -> 5Fe3+ + Mn2+ + 4H2O
From the stoichiometry of the reaction, we know that 1 mole of KMnO4 reacts with 5 moles of Fe2+. Therefore, the number of moles of Fe2+ that reacted can be calculated as:
moles of Fe2+ = (0.004053 moles KMnO4) × (5 moles Fe2+/1 mole KMnO4) = 0.020264 moles
Now, we need to calculate the molar mass of Fe. The molar mass of Fe is 55.845 g/mol.
mass of Fe = (moles of Fe2+) × (molar mass of Fe)
= 0.020264 moles × 55.845 g/mol
= 1.1326 grams
Finally, we can calculate the percentage of Fe in the iron ore sample by dividing the mass of Fe by the mass of the sample and multiplying by 100:
percentage of Fe = (mass of Fe / mass of sample) × 100
= (1.1326 grams / 0.446 grams) × 100
= 253.80%
Therefore, the percentage of Fe in the iron ore sample is approximately 253.80%.