Respuesta :
For Lead (II) Nitrate + Potassium Iodide → Lead (II) Iodide + Potassium Nitrate , The balanced equation is [tex]Pb(NO_{3})_{2}[/tex] + 2KI -----> [tex]PbI_{2}[/tex] + 2 [tex]KNO_{3}[/tex]
We must first convert from a word equation to a symbol equation:
Lead (II) Nitrate + Potassium Iodide → Lead (II) Iodide + Potassium
Nitrate
The lead (II) ion is represented as [tex]Pb^{2+}[/tex] and nitrate ion as [tex]NO_{3} ^{-}[/tex]
We need two nitrate ions per lead (II) ion, so lead (II) nitrate is [tex]Pb(NO_{3})_{2}[/tex]
The potassium iodide is simply KI
In lead (II) iodide, the charges balance in a 1:2 ratio, so it is [tex]PbI_{2}[/tex]
And, potassium nitrate is simply [tex]KNO_{3}[/tex]
So, The symbol equation is as follows:
[tex]Pb(NO_{3})_{2}[/tex] + KI -----> [tex]PbI_{2}[/tex] + [tex]KNO_{3}[/tex]
Now, increase the number of nitrate ions on the right hand side of the equation as:
[tex]Pb(NO_{3})_{2}[/tex] + KI -----> [tex]PbI_{2}[/tex] + 2 [tex]KNO_{3}[/tex]
Now, balance the potassium ions on each side of the equation as:
[tex]Pb(NO_{3})_{2}[/tex] + 2KI -----> [tex]PbI_{2}[/tex] + 2 [tex]KNO_{3}[/tex]
Hence, the balanced equation is ;
[tex]Pb(NO_{3})_{2}[/tex] + 2KI -----> [tex]PbI_{2}[/tex] + 2 [tex]KNO_{3}[/tex]
Learn more about potassium iodide here:
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