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 `K_(2) [PtCl_(4)]` and `NH_(3)` used, respectively, are

Options

A. 0.1,0.2
B. 0.2,0.4
C. 0.3,0.6
D. 0.03,0.06

Explanation

To solve the problem, we need to calculate the number of moles of \( K_2[PtCl_4] \) and \( NH_3 \) given their respective masses and atomic weights.

### Step 1: Calculate the molar mass of \( K_2[PtCl_4] \)

The formula for \( K_2[PtCl_4] \) consists of:
- 2 Potassium (K)
- 1 Platinum (Pt)
- 4 Chlorine (Cl)

Using the atomic weights provided:
- Atomic weight of K = 39 g/mol
- Atomic weight of Pt = 415 g/mol
- Atomic weight of Cl = 35.5 g/mol

Calculating the molar mass:
\[ \text{Molar mass of } K_2[PtCl_4] = (2 \times 39) + (1 \times 415) + (4 \times 35.5) \]
\[ = 78 + 415 + 142 = 635 \text{ g/mol} \]

### Step 2: Calculate the number of moles of \( K_2[PtCl_4] \)

Given mass of \( K_2[PtCl_4] = 83.0 \text{ g} \)

Using the formula for moles:
\[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} \]
\[ = \frac{83.0 \text{ g}}{635 \text{ g/mol}} \approx 0.130 \text{ moles} \]

### Step 3: Calculate the molar mass of \( NH_3 \)

The formula for \( NH_3 \) consists of:
- 1 Nitrogen (N)
- 3 Hydrogen (H)

Using the atomic weights provided:
- Atomic weight of N = 14 g/mol
- Atomic weight of H = 1 g/mol

Calculating the molar mass:
\[ \text{Molar mass of } NH_3 = (1 \times 14) + (3 \times 1) = 14 + 3 = 17 \text{ g/mol} \]

### Step 4: Calculate the number of moles of \( NH_3 \)

Given mass of \( NH_3 = 83.0 \text{ g} \)

Using the formula for moles:
\[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} \]
\[ = \frac{83.0 \text{ g}}{17 \text{ g/mol}} \approx 4.88 \text{ moles} \]

### Summary of Results
- Number of moles of \( K_2[PtCl_4] \): \( \approx 0.130 \) moles
- Number of moles of \( NH_3 \): \( \approx 4.88 \) moles