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]`
Which reactant is the limiting reagent and which is in excess?

Options

A. `K_(2)[PtCl_(4)]` `NH_(3)`
B. `NH_(3)` `K_(2) [PtCl_(4)]`
C. None None
D. Both Both

Explanation

To determine the limiting reagent and the excess reagent in the reaction between potassium tetra-chloroplatinate (K₂[PtCl₄]) and ammonia (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 (Potassium) = 39 g/mol
- Pt (Platinum) = 415 g/mol
- Cl (Chlorine) = 35.5 g/mol

The formula for K₂[PtCl₄] consists of:
- 2 K: \(2 \times 39 = 78\) g/mol
- 1 Pt: \(1 \times 415 = 415\) g/mol
- 4 Cl: \(4 \times 35.5 = 142\) g/mol

Now, adding these together:
\[ \text{Molar mass of K₂[PtCl₄]} = 78 + 415 + 142 = 635 \text{ g/mol} \]

### Step 2: Calculate the molar mass of NH₃
Using the atomic weights:
- N (Nitrogen) = 14 g/mol
- H (Hydrogen) = 1 g/mol

The formula for NH₃ consists of:
- 1 N: \(1 \times 14 = 14\) g/mol
- 3 H: \(3 \times 1 = 3\) g/mol

Now, adding these together:
\[ \text{Molar mass of NH₃} = 14 + 3 = 17 \text{ g/mol} \]

### Step 3: Calculate the number of moles of each reactant
Using the formula:
\[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} \]

For K₂[PtCl₄]:
\[ \text{Moles of K₂[PtCl₄]} = \frac{83 \text{ g}}{635 \text{ g/mol}} \approx 0.130 \text{ moles} \]

For NH₃:
\[ \text{Moles of NH₃} = \frac{83 \text{ g}}{17 \text{ g/mol}} \approx 4.88 \text{ moles} \]

### Step 4: Determine the stoichiometry of the reaction
The balanced chemical equation is:
\[ \text{K₂[PtCl₄]} + 2 \text{NH₃} \rightarrow \text{Cisplatin} + 2 \text{KCl} \]

From the equation, we see that:
- 1 mole of K₂[PtCl₄] reacts with 2 moles of NH₃.

### Step 5: Calculate the required moles of NH₃ for the available K₂[PtCl₄]
Since we have 0.130 moles of K₂[PtCl₄], we need:
\[ \text{Required moles of NH₃} = 0.130 \text{ moles K₂[PtCl₄]} \times 2 = 0.260 \text{ moles NH₃} \]

### Step 6: Compare the available moles of NH₃ with the required moles
We have approximately 4.88 moles of NH₃ available, which is much greater than the 0.260 moles required.

### Conclusion
- **Limiting Reagent**: K₂[PtCl₄] (as it will be consumed first)
- **Excess Reagent**: NH₃ (as there is more than enough available)