Question

Cisplatin Pt(NH3)2Cl2 has been studied and used as an anti-tumor agent to treat certain types of cancer. The compound is generated by the following reaction:K2PtCl4 2NH3 -> Pt(NH3)2Cl2 2KCl A pharmacy supplier combines 655.1 Kg K2PtCl4 with 728 Kg NH3 to make cis-plat for an order from a hospital chain. a. Determine the limiting and excess reagent.b. What is the maximum amount of cis-plat produced

Answer 1

* Limiting reactant: K2PtCl4.

*
`m_(Pt(NH_3)_2Cl_2)=475kgPt(NH_3)_2Cl_2`

Step-by-step explanation:

Hello,

In this case, for the reaction:

`K_2PtCl_4 +2NH_3 \rightarrow Pt(NH_3)_2Cl_2 +2KCl`

By starting with 655.1 kg of K2PtCl4 (molar mass 415.1 g/mol) and 728 kg of NH3 (molar mass 17 g/mol) the limiting reactant is identified as the one yielding the smallest moles of cisplatin, thus, we proceed as follows:

`n_(Pt(NH_3)_2Cl_2)^(by \ K_2PtCl_4)=655.1kg K_2PtCl_4*(1kmol K_2PtCl_4)/(415.1g K_2PtCl_4)*(1molPt(NH_3)_2Cl_2)/(1mol K_2PtCl_4) =1.578kmolPt(NH_3)_2Cl_2\\\\n_(Pt(NH_3)_2Cl_2)^(by \ NH_3)=728kgNH_3*(1kmol NH_3)/(17g NH_3)*(1molPt(NH_3)_2Cl_2)/(2molNH_3) =21.4kmolPt(NH_3)_2Cl_2`

In such a way, we infer that the K2PtCl4 is the limiting reactant, therefore, the maximum amount of cis-plat (300 g/mol) produced is:

`m_(Pt(NH_3)_2Cl_2 )=1.578kmolPt(NH_3)_2Cl_2*(301gPt(NH_3)_2Cl_2)/(1molPt(NH_3)_2Cl_2)\\ \\m_(Pt(NH_3)_2Cl_2)=475kgPt(NH_3)_2Cl_2`

Best regards.