A solution is prepared by adding 0.063 mole of K3 [Fe(CN)6] to 0.59 L of 2.1...

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Question

A solution is prepared by adding 0.063 mole of K3 [Fe(CN)6] to 0.59 L of 2.1 M NaCN. Assuming no volume change, calculate the

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Answer 1

Answer #1

The detailed steps are given below

Quitially, [Na CN] = 2.1M (wsfecci) | < 0.063 male = 0-10678 = 0.063 mole o.Sa L 0.10678 Fe²+ + GCN [Fe(CN)6] . H 2-1 C + x +

Answer 2

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