A 0.200g sample containing mn02 was dissolved and analyzed by addiction of 50ml of 0.100M fe to drive the reation. The excess fe required 15ml of 0.0200 m kmn04. Find % mn304 fw 228.8mg/mmol in the sample?
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I can help you with that! To find the percentage of MnO4 in the sample, we need to calculate the amount of MnO4 consumed by Fe and then determine the percentage of MnO4 in the original sample.
First, we need to determine the number of moles of Fe that reacted with the MnO2.
Given:
Mass of the MnO2 sample = 0.200 g
Concentration of Fe solution = 0.100 M
Volume of Fe solution used = 50 mL = 0.050 L
To find the number of moles of Fe, we will use the equation:
moles of Fe = concentration of Fe × volume of Fe solution
= 0.100 M × 0.050 L
Next, we need to determine the number of moles of MnO4 reduced by Fe.
Given:
Volume of KMnO4 solution used = 15 mL = 0.015 L
Concentration of KMnO4 solution = 0.0200 M
Using the equation:
moles of KMnO4 = concentration of KMnO4 × volume of KMnO4 solution
= 0.0200 M × 0.015 L
Since MnO2 and MnO4 react in a 1:1 stoichiometric ratio, the moles of MnO4 reduced by Fe are equal to the moles of Fe consumed:
moles of MnO4 reduced = moles of Fe
Now, we can calculate the mass of MnO4 reduced:
mass of MnO4 reduced = moles of MnO4 reduced × molar mass of MnO4
Given that the molar mass of MnO4 is 158.034 g/mol, you can calculate the mass of MnO4 reduced.
Finally, to find the percentage of MnO4 in the sample, we divide the mass of MnO4 reduced by the mass of the sample and multiply by 100:
percentage of MnO4 in sample = (mass of MnO4 reduced / mass of sample) × 100
Given that the molar mass of MnO4 is 228.8 mg/mmol, you can convert the mass of MnO4 reduced to moles and then calculate the percentage of MnO4 in the sample.
Please note that the % MnO4 calculated assumes that all the MnO2 in the sample reacted with Fe and was converted to MnO4.