Question
A mixture of 3.00 g of silver and copper metals was dissolved in excess nitric acid.
Resulting salts, silver(I) nitrate and copper(II) nitrate, were isolated and dissolved in enough
water to make 0.100 L of a solution with a total concentration of the nitrate ion equal to 0.650 M.
Calculate the mass percentage of silver metal in the starting mixture.
Resulting salts, silver(I) nitrate and copper(II) nitrate, were isolated and dissolved in enough
water to make 0.100 L of a solution with a total concentration of the nitrate ion equal to 0.650 M.
Calculate the mass percentage of silver metal in the starting mixture.
Answers
DrBob222
This is a problem with two equations and two unknowns and they are solved simultaneously.
Let X = mass Ag
and Y = mass Cu
-----------------
equation 1 is X + Y = 3.00
To get the second equation note that mols NO3^- = M x L = 0.650 M x 0.1L = 0.0650 mols. Therefore, mols NO3^- from Ag + mols NO3^- from Cu = 0.0650. To save typing I'll let am stand for atomic mass. Therefore, equation 2 is
(X/am Ag) + (2Y/am Cu) = 0.0650
Solve those two equations for X and Y simultaneously, then
%Ag = (mass Ag/mass sample)* 100 = (X/3.0)*100 = % Ag (mass percent)
Sol
Let X = mass Ag
and Y = mass Cu
-----------------
equation 1 is X + Y = 3.00
To get the second equation note that mols NO3^- = M x L = 0.650 M x 0.1L = 0.0650 mols. Therefore, mols NO3^- from Ag + mols NO3^- from Cu = 0.0650. To save typing I'll let am stand for atomic mass. Therefore, equation 2 is
(X/am Ag) + (2Y/am Cu) = 0.0650
Solve those two equations for X and Y simultaneously, then
%Ag = (mass Ag/mass sample)* 100 = (X/3.0)*100 = % Ag (mass percent)
Sol