Copper sulphate reacts with NaCN to form a cyanide complex. Write the balanced equation and find the number of NaCN molecules involved in the equation for one mole of `CuSO_(4)`.

Explanation

To solve the problem of how copper sulfate reacts with sodium cyanide (NaCN) to form a cyanide complex, we will follow these steps:

### Step 1: Write the initial reaction
Copper sulfate (CuSO₄) reacts with sodium cyanide (NaCN) to form cupric cyanide (CuCN) and sodium sulfate (Na₂SO₄). The unbalanced equation can be written as:
\[ \text{CuSO}_4 + \text{NaCN} \rightarrow \text{CuCN} + \text{Na}_2\text{SO}_4 \]

### Step 2: Balance the first reaction
To balance the first reaction, we note that one mole of CuSO₄ produces one mole of Na₂SO₄, but we need to balance the sodium (Na) atoms. Since Na₂SO₄ has 2 Na, we need 2 NaCN:
\[ \text{CuSO}_4 + 2 \text{NaCN} \rightarrow \text{CuCN} + \text{Na}_2\text{SO}_4 \]

### Step 3: Consider the stability of CuCN
Cupric cyanide (CuCN) is not stable and can react further with NaCN to form a complex. The reaction can be represented as:
\[ \text{CuCN} + 3 \text{NaCN} \rightarrow \text{Na}_3\text{Cu(CN)}_4 \]

### Step 4: Combine the reactions
Now we need to combine the balanced reactions. From the first reaction, we have:
\[ \text{CuSO}_4 + 2 \text{NaCN} \rightarrow \text{CuCN} + \text{Na}_2\text{SO}_4 \]
And from the second reaction:
\[ \text{CuCN} + 3 \text{NaCN} \rightarrow \text{Na}_3\text{Cu(CN)}_4 \]

### Step 5: Balance the overall reaction
To balance the overall reaction, we need to account for all the sodium cyanide used. The total NaCN used is:
- 2 NaCN from the first reaction
- 3 NaCN from the second reaction

Thus, for one mole of CuSO₄, we have:
\[ 2 \text{NaCN} + 3 \text{NaCN} = 5 \text{NaCN} \]

### Final Balanced Equation
The overall balanced equation combining both reactions is:
\[ \text{CuSO}_4 + 5 \text{NaCN} \rightarrow \text{Na}_3\text{Cu(CN)}_4 + \text{Na}_2\text{SO}_4 \]

### Conclusion
For one mole of CuSO₄, **5 moles of NaCN** are required.

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