Question

The amount of Fe2 in 2.19 g of sample is determined by titrating it with MnO4 minus. If 25.0 ml of 0.1000M MnO4 minus is used to get to the endpoint, calculate the percent mass.

Answer 1

To find the percent mass of Fe₂+ in the sample, calculate the moles of MnO₄- used, use stoichiometry to find the moles of Fe₂+, convert that to grams, and then calculate the percent mass based on the sample's mass.

Step-by-step explanation:

To calculate the percent mass of Fe₂+ based on the titration with MnO₄-, you first need to identify the reaction and use stoichiometry to determine the amount of Fe₂+ that corresponds to the amount of MnO₄- used. For the titration of Fe₂+ with MnO₄-, the balanced chemical equation is typically:

5 Fe₂+ + MnO₄- + 8 H+ → 5 Fe₃+ + Mn²+ + 4 H₂O

Based on the stoichiometry of the reaction, 1 mole of MnO₄- reacts with 5 moles of Fe₂+. Knowing that 25.0 mL of 0.1000 M MnO₄- is used, you can calculate the moles of MnO₄- and subsequently the moles of Fe₂+:

• Moles of MnO₄- = 0.1000 M * 0.0250 L = 0.0025 moles
• Moles of Fe₂+ = 0.0025 moles MnO₄- * (5 moles Fe₂+/1 mole MnO₄-) = 0.0125 moles Fe₂+

Now, convert the moles of Fe₂+ to grams using the molar mass of Fe, which is approximately 55.85 g/mol:

Grams of Fe₂+ = 0.0125 moles * 55.85 g/mol = 0.6981 g

Finally, the percent mass of Fe₂+ in the sample is calculated by dividing the mass of iron by the mass of the sample and multiplying by 100:

Percent mass = (0.6981 g Fe₂+ / 2.19 g sample) * 100 = 31.89%