Question

The compound cisplatin has been extensively studied as an antitumor agent. It is synthesized by the following, unbalanced reaction. What mass of cisplatin can be made from 65g of K2PtCl4 with sufficient NH3?
K2PtCl4(aq) + NH3(aq) --> Pt(NH3)2Cl2(s) + KCl(aq)

Answer 1

Answer:

46.98 grams of cisplatin can be made from 65 grams of potassium tetrachloridoplatinate(II).

Explanation:

$K_2PtCl_4(aq) + 2NH_3(aq)\rightarrow Pt(NH_3)_2Cl_2(s) + 2KCl(aq)$

Mass of potassium tetrachloridoplatinate(II) = 65 g

Moles of potassium tetrachloridoplatinate(II) = $\frac{65}{415 g/mol}=0.1566 mol$

1 mol of potassium tetrachloridoplatinate(II) gives 1 mol of cisplatin.

Then 0.1566 moles of potassium tetrachloridoplatinate(II) will give:

$\frac{1}{1}\times 0.1566 mol=0.1566 mol$

Mass of 0.1566 moles of cisplatin:

$0.1566 mol\times 300 g/mol=46.98 g$

46.98 grams of cisplatin can be made from 65 grams of potassium tetrachloridoplatinate(II).

Answer 2

The balanced chemical equation would be as follows:

K2PtCl4(aq) + 2NH3(aq) --> Pt(NH3)2Cl2(s) + 2KCl(aq)

We are given the amount of K2PtCl4 to be used in the reaction. This will be the starting point for our calculations. We do as follows:

65 g K2PtCl4 ( 1 mol / 415.09 g ) ( 1 mol Pt(NH3)2Cl2 / 1 mol K2PtCl ) ( 300.051 g / 1 mol ) = 46.99 g Pt(NH3)2Cl produced