Question

Consider the reaction of 19.0 g of zinc metal with excess silver nitrate to produce silver metal and zinc nitrate. The reaction is stopped before all the zinc metal has reacted and 29.0 g of solid metal is present. If 14.6 g of zinc are present in the solid metal, calculate the mass of silver metal present in the mixture.

Answer 1

Answer:

14.5 g silver

Explanation:

This is a problem using the stoichiometry of the reaction. First thing we need is the balanced equation:

   Zn + 2 AgNO3 ----------------------- 2 Ag + Zn(NO3)2

We know that 14.6 g of Zn did not reacted, then we can calculate the amount of Zn reacted and do the calculation given the above reaction.

amount Zn reacted: 19.0 -14.6 g Zn = 4.4 g Zn

atomic weight of Zn:  65.37 g/mol

mol Zn reacted: 4.4 g Zn x ( 1 mol Zn/ 65.37 g Zn) = 0.067 mol Zn

We know from the balanced equation that moles of Ag are produced from 1 mol Zn therefore the mol of Ag produced are:

0.067 mol Zn x 2 mol Ag/ 1mol Zn = 0.135 mol Ag

and the mass of silver then will be given by multiplying by the atomic weight of silver:

0.135 mol Ag x  107.9 g/mol = 14.5 g Ag