Silver is removed from the solutions of its salts with metallic zinc, according to the reaction
`Zn + 2 Ag^(o+) rarr Zn^(2+) + 2Ag`.
A `65.4 g` piece of `Zn` is put into a `100 L` vat containing `3.25 g` `Ag^(o+)` per litre. How amny moles of reactant remained unreacted?

Answer

A

Explanation

Total Weight of `Ag^(o+) = ((3.24 g)/(1L)) (100 L) = 324 g`
Mol of `Ag^(o+) = (324)/(108 g Ag) = 3 "mol"`
Mol of `Zn = (65.4)/(65.4 g Zn) = 1 "mol" = 2 "mol of" Ag^(o+)`
Since the mole ratio `Ag^(o+)//Zn` present `(3//1)` exceeds the mole ratio required `(2//1)` in the reaction, the `Ag^(o+)` is in excess, the `Zn` is completely consumed.
Moles of `Ag^(o+)` consumed = 2 mol
Mole of `Ag^(o+)` left `= 3 - 2 = 1` mol of `Ag^(o+)` excess