Silver metal can be prepared by reducing its nitrate, AgNO3 with copper according to the following equation:
1 answer:
%yield = 88.5%
<h3>Further explanation</h3>
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%
<em>%yield = 88.5%</em>
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