Cisplation is used an anticancer agent for the treatment of solid tumors, and its prepared as follows:
`underset({:("Potassium tetra"),("Chloro platinate"):})(K_(2)[PtCl_(4)]) + 2NH_(3) rarr underset("Ciplatin")([Pt(NH_(3))_(2) Cl_(2)]) + 2KCl`
Given `83.0 g` of `K_(2) [PtCl_(4)]` is used with `83.0 g` of `NH_(3)`.
Atomic weights: `K = 39 , Pt = 415, Cl = 35.5 N = 14]`
The number of mol of excess reactant is

Options

A. 4.68
B. 4.78
C. 4.58
D. 4.48

Explanation

To solve the problem, we need to determine the number of moles of the excess reactant when 83.0 g of K₂[PtCl₄] reacts with 83.0 g of NH₃. We will follow these steps:

### Step 1: Calculate the molar mass of K₂[PtCl₄]
The molar mass can be calculated using the atomic weights provided:
- K = 39 g/mol
- Pt = 415 g/mol
- Cl = 35.5 g/mol

The formula for K₂[PtCl₄] is:
\[ \text{Molar mass of K₂[PtCl₄]} = 2 \times \text{K} + \text{Pt} + 4 \times \text{Cl} \]
\[ = 2 \times 39 + 415 + 4 \times 35.5 \]
\[ = 78 + 415 + 142 \]
\[ = 635 \text{ g/mol} \]

### Step 2: Calculate the number of moles of K₂[PtCl₄]
Using the formula:
\[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} \]
\[ = \frac{83.0 \text{ g}}{635 \text{ g/mol}} \]
\[ \approx 0.130 \text{ moles} \]

### Step 3: Calculate the molar mass of NH₃
The molar mass of NH₃ is calculated as follows:
- N = 14 g/mol
- H = 1 g/mol

The formula for NH₃ is:
\[ \text{Molar mass of NH₃} = \text{N} + 3 \times \text{H} \]
\[ = 14 + 3 \times 1 \]
\[ = 14 + 3 \]
\[ = 17 \text{ g/mol} \]

### Step 4: Calculate the number of moles of NH₃
Using the formula:
\[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} \]
\[ = \frac{83.0 \text{ g}}{17 \text{ g/mol}} \]
\[ \approx 4.88 \text{ moles} \]

### Step 5: Determine the limiting reagent
From the balanced equation:
\[ \text{K₂[PtCl₄]} + 2 \text{NH₃} \rightarrow \text{Cisplatin} + 2 \text{KCl} \]

This shows that 1 mole of K₂[PtCl₄] reacts with 2 moles of NH₃.

Now, we need to find out how many moles of NH₃ are required for the available moles of K₂[PtCl₄]:
- For 0.130 moles of K₂[PtCl₄], the required moles of NH₃:
\[ 0.130 \text{ moles K₂[PtCl₄]} \times 2 = 0.260 \text{ moles NH₃} \]

### Step 6: Identify the limiting reagent
We have:
- Available moles of K₂[PtCl₄] = 0.130 moles
- Available moles of NH₃ = 4.88 moles
- Required moles of NH₃ = 0.260 moles

Since we have enough NH₃ to react with K₂[PtCl₄], K₂[PtCl₄] is the limiting reagent.

### Step 7: Calculate the excess moles of NH₃
Now we can find the moles of NH₃ that reacted:
- Moles of NH₃ that reacted = 0.130 moles K₂[PtCl₄] × 2 = 0.260 moles NH₃

Now, we can find the excess moles of NH₃:
\[ \text{Excess moles of NH₃} = \text{Initial moles of NH₃} - \text{Moles of NH₃ that reacted} \]
\[ = 4.88 - 0.260 \]
\[ = 4.62 \text{ moles} \]

### Final Answer
The number of moles of the excess reactant (NH₃) is approximately **4.62 moles**.

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