Question

4) Cisplatin, Pt(NH3)2Cl2, has been studied as an antitumor agent. It is synthesized as follows:
K PtClafaq) + 2NH3(aq) → Pt(NH3)2Cl2(s)
+ 2 KClag)
What mass of cisplatin can be produced from 100. g of KzPtCle and excess ammonia?

Answer 1

Answer:

Mass of cisplatin =   72.32 g

Explanation:

Given data:

Mass of KPtCl = 100 g

Mass of Cisplatin = ?

Solution:

First of all we will write the balance chemical equation.

Chemical equation:

K₂PtCl₄ (aq) + 2NH₃(aq)  → Pt(NH₃)₂Cl₂(s)  +  2KCl (aq)

Number of moles of = mass / molar mass

Molar mass of K₂PtCl₄ = 39.0983 ×2 + 195.084 + 35.5× 4

Molar mass of K₂PtCl₄ = 415.284 g/mol

Number of moles of K₂PtCl₄ = 100 g / 415.284 g/mol

Number of moles of K₂PtCl₄ = 0.241 mol

Now we will compare the moles of K₂PtCl₄ with Cisplatin from balance chemical equation.

                  K₂PtCl₄           :         Pt(NH₃)₂Cl₂

                       1                 :               1

                      0.241           :              0.241

Ammonia is in excess so mass of cisplatin depend upon the K₂PtCl₄.

Mass of cisplatin:

Mass of cisplatin = Number of moles × Molar mass

Mass of cisplatin =   0.241 mol × 300.1 g/mol

Mass of cisplatin =   72.32 g