Silver metal can be prepared by reducing its nitrate, AgNO3 with copper according to the following equation: Cu(s) + 2 AgNO3(aq) → Cu(NO3)2(aq) + 2 Ag(s)

What is the percent yield of the reaction if 71.5 grams of Ag was obtained from 127.0 grams of AgNO3 ?

%yield = 88.5%

Given

Reaction

Cu(s) + 2 AgNO₃(aq) → Cu(NO₃)₂(aq) + 2Ag(s)

Required

The percent yield

Solution

mol AgNO₃(MW=169,87 g/mol) :

= mass : MW

= 127 : 169.87

= 0.748

mol Ag from equation :

= 2/2 x mol AgNO₃

= 2/2 x 0.748

= 0.748

Mass Ag (theoretical) :

= mol x Ar Ag

= 0.748 x 108

= 80.784

% yield = (actual/theoretical) x 100%

%yield = 71.5/80.784 x 100%

%yield = 88.5%

the answer is a! (2.9 x 10^-18 j) hope i could someone. : )