Answer to Question #143577 in General Chemistry for unknowns

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