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Au+CN^(-)+H(2)+O(2) to [Au(CN)(2)+OH]^(-...

`Au+CN^(-)+H_(2)+O_(2) to [Au(CN)_(2)+OH]^(-)`. How many `CN^(-)` ions are involved in the above balanced equation? (per mole of Au)

Text Solution

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The correct Answer is:
2
`4Au+8NaCN+2H_(2)O+O_(2) to 4Na[Au(CN)_(2)]+4NaOH`.
4 moles of Au..... 8 moles of NaCn
1 mole...... ? `:.` 2 ole of NaCN is used for 1 mole of .Au.
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