How many cyanide ions are involved in the following chemical equation ?
`Au + CN^(Ө)+ H_(2)O +O_(2) rarr [Au(CN)_(2)]^(Ө) + overset(Ө) O H`.

Explanation

To determine how many cyanide ions are involved in the given chemical equation, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants and Products**:
The equation provided is:
\[ Au + CN^{-} + H_{2}O + O_{2} \rightarrow [Au(CN)_{2}]^{-} + OH^{-} \]
Here, the reactants include gold (Au), cyanide ions (CN^-), water (H2O), and oxygen (O2). The products include the complex ion [Au(CN)2]^- and hydroxide ions (OH^-).

2. **Balance the Equation**:
To find the number of cyanide ions, we need to balance the equation. According to the video transcript, the balanced equation is:
\[ 4Au + 8CN^{-} + 4H_{2}O + O_{2} \rightarrow 4[Au(CN)_{2}]^{-} + 4OH^{-} \]
This indicates that for every 4 moles of gold, 8 moles of cyanide ions are required.

3. **Count the Cyanide Ions**:
From the balanced equation, we see that 8 moles of cyanide ions (CN^-) are involved in the reaction.

4. **Final Answer**:
Therefore, the number of cyanide ions involved in the reaction is **8 moles**.

### Summary:
The total number of cyanide ions involved in the reaction is **8 moles**.

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