Question

The compound cisplatin, Pt(NH3)2Cl2, has been used as an antitumor agent. It is prepared by the reaction between potassium tetrachloroplatinate (K2PtCl4) and ammonia (NH3).

K2PtCl4 + 2 NH3 → Pt(NH3)2Cl2 + 2 KCl

What mass (in grams) of Pt(NH3)2Cl2 can be obtained starting with 0.8862 g K2PtCl4?

Answer 1

The equation for the reaction is;

$K_2PtCl_4+2NH_3\to Pt(NH_3)_2Cl_2+2KCl$

$1mol$ of $K_2PtCl_4$ yields $1mol of Pt(NH_3)_2Cl_2$

Mass of $K_2PtCl_4$ $=0.8862g$

Mass of $Pt(NH_3)_2Cl_2$ $= unknown$

Molar mass of $K_2PtCl_4=415g/mol$

Molar mass of $Pt(NH_3)_2Cl_2=301.1g/mol$

Number of moles of $K_2PtCl_4=$ $0.8862g\over 415g/mol$ $=0.00214moles$

Mass of

$Pt(NH_3)_2Cl_2$ $=0.00214mol K_2Pt_Cl_4\times$ $301.1g/mol [Pt(NH_3)_2Cl_2]\over 1mol [Pt(NH_3)_2Cl_2]$ $=0.644g$

Mass of $[Pt(NH_3)_2Cl_2]$ $=0.644g$