10-2 mole of Fe3O4 is treated with excess KI solution in presence of dilute H2SO4, the products are Fe2+ and I2(g). What volume of 0.1 (M) Na2S2O3 will be needed to reduce the liberated I2(g)?
Correct answer: C. 200 ml
- A. 50 ml
- B. 100 ml
- C. 200 ml
- D. 400 ml
Explanation
When Fe3O4 reacts with excess KI and dilute H2SO4, it produces Fe2+ and I2. For every mole of Fe3O4, 4 moles of I2 are produced. Given 10-2 moles of Fe3O4, 4 x 10-2 moles of I2 are produced. The reaction of I2 with Na2S2O3 is a 1:2 stoichiometry. Thus, 8 x 10-2 moles of Na2S2O3 are needed. With a 0.1 M solution, the volume required is 200 ml. Hence, Option C is correct. Options A, B, and D are incorrect due to miscalculations of the stoichiometric balance and required volume.
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