Respuesta :
Answer:
[tex]Cu~+~PtCl_2->Pt~+~CuCl_2[/tex]
Explanation:
In this case, we can start with the formula of Platinum (II) Chloride. The cation is the atom at the left of the name (in this case [tex]Pt^+^2[/tex]) and the anion is the atom at the right of the name (in this case [tex]Cl^-[/tex]). With this in mind, the formula would be [tex]PtCl_2[/tex].
Now, if we used metallic copper we have to put in the reaction only the copper atom symbol [tex]Cu[/tex]. So, we have as reagents:
[tex]Cu~+~PtCl_2->[/tex]
The question now is: What would be the products? To answer this, we have to remember "single displacement reactions". With a general reaction:
[tex]A~+~BC->AB~+~C[/tex]
With this in mind, the reaction would be:
[tex]Cu~+~PtCl_2->Pt~+~CuCl_2[/tex]
I hope it helps!
