Answer:
* Limiting reactant: K2PtCl4.
* [tex]m_{Pt(NH_3)_2Cl_2}=475kgPt(NH_3)_2Cl_2[/tex]
Explanation:
Hello,
In this case, for the reaction:
[tex]K_2PtCl_4 +2NH_3 \rightarrow Pt(NH_3)_2Cl_2 +2KCl[/tex]
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:
[tex]n_{Pt(NH_3)_2Cl_2}^{by \ K_2PtCl_4}=655.1kg K_2PtCl_4*\frac{1kmol K_2PtCl_4}{415.1g K_2PtCl_4}*\frac{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*\frac{1kmol NH_3}{17g NH_3}*\frac{1molPt(NH_3)_2Cl_2}{2molNH_3} =21.4kmolPt(NH_3)_2Cl_2[/tex]
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:
[tex]m_{Pt(NH_3)_2Cl_2 }=1.578kmolPt(NH_3)_2Cl_2*\frac{301gPt(NH_3)_2Cl_2}{1molPt(NH_3)_2Cl_2}\\ \\m_{Pt(NH_3)_2Cl_2}=475kgPt(NH_3)_2Cl_2[/tex]
Best regards.
