Au+CN^(-)+H_(2)+O_(2) to [Au(CN)_(2)+OH]^(-). How many CN^(-) ions are – InterviewSolution

1.	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)
Answer» Solution :`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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