Question

4.9 gr K2Cr2O7 is taken to prepare 0.1 litre of the solution. 10 ml of this solution is further taken to oxidise SN^+2 ion into Sn^+4 ion. the SN+4 so produced is used in 2nd reaction to prepare Fe^+3 ion, then the millimoles of Fe^+3 ion formed will be (assume all other components are in sufficient amounts)

Answer 1

Answer on the question #54222 – Chemistry – Physical Chemistry

Question:

4.9 gr K₂Cr₂O₇ is taken to prepare 0.1 litre of the solution. 10 ml of this solution is further taken to oxidise SN⁺² ion into Sn⁺⁴ ion. The SN⁺⁴ so produced is used in 2nd reaction to prepare Fe⁺³ ion, then the millimoles of Fe⁺³ ion formed will be (assume all other components are in sufficient amounts).

Solution:

The equations of chemical reactions are:

$$\begin{array}{l} Cr_2O_7^{2-} + 3Sn^{2+} + 14H^+ = 2Cr^{3+} + 3Sn^{4+} + 7H_2O, \\ Sn^{4+} + 2Fe^{2+} = Sn^{2+} + 2Fe^{3+}. \end{array}$$

According to these equations, the number of moles of Cr₂O₇²⁻ anions and number of moles of Fe³⁺ ions relate as:

$$\begin{array}{l} n(Cr_2O_7^{2-}) = \frac{n(Sn^{4+})}{3}, \\ n(Sn^{4+}) = \frac{n(Fe^{3+})}{2}, \\ n(Fe^{3+}) = 2n(Sn^{4+}) = 2 \times 3 \times n(Cr_2O_7^{2-}). \end{array}$$

Number of moles of Cr₂O₇²⁻ can be calculated from the solution preparation data:

$$n(Cr_2O_7^{2-}) = \frac{m(K_2Cr_2O_7^{2-})}{M(K_2Cr_2O_7^{2-})} \times \frac{V(\text{sample})}{V(\text{total})} = \frac{4.9}{294.185} \times \frac{0.01}{0.1} = 1.666 \text{ mmol}.$$

Then, the number of moles of Fe³⁺ is:

$$n(Fe^{3+}) = 6 \times n(Cr_2O_7^{2-}) = 9.994 \text{ mmol}.$$

Answer: 9.994 mmol (millimoles) of Fe³⁺ is formed.

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