Home
Class 11
CHEMISTRY
Silver is removed from the solutions of ...

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?

Text Solution

Verified by Experts

The correct Answer is:
A

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
Promotional Banner

Similar Questions

Explore conceptually related problems

Write the electrochemical cell for the overall cell reaction Zn_((s))+2AgNO_(3)rarr2Ag_((s))+Zn(NO_(3))_(2) .

How much magnesium sulphide can be obtained from 2.00 g of Mg and 2.00 g of S by the reaction. Mg + S rarr MgS . Which is the limiting reagent? Calculate the amount of one of the reactants which remains unreacted?

For the reaction : C_(3)H_(8)(g) + 5O_(2) rarr 3CO_(2)(g) + 4H_(2)O(l) a constant temperature, Delta H - Delta U is

At constant temperature in a litre vessel , when the reaction 2SO_(3)(g) iff 2SO_(2)(g) + O_(2)(g) is 2.25 litre "mol"^(-1) . What would be the concentration of Y at equilibrium with 2.0 moles of X and 3.0 moles of Z in one litre vessel ?

Calculate the emf of the cell having the cell reaction 2Ag^(+)+ZnhArr 2Ag+Zn^(2+) and E_("cell")^(@)=1.56V" at "25^(@)C when concentration of Zn^(2+)=0.1M and Ag^(+)=10M in the solution.