# Given the equation: Cu + AgNO3 --> Ag + CuNO3, what mass of silver in grams is precipitated when 40.0 g of copper reacts with an excess of silver nitrate in solution?

Dec 4, 2015

$68.0 g$ of silver.

#### Explanation:

The reaction is:

$C u \left(s\right) + A g N {O}_{3} \left(a q\right) \to A g \left(s\right) + C u N {O}_{3} \left(a q\right)$

Since the silver nitrate is in excess, then copper is the limiting reactant , and therefore, its mass will be used to calculate the mass of silver that will form:

?gAg=40.0cancel(gCu)xx(cancel(1molCu))/(63.5cancel(gCu))xx(1cancel(molAg))/(1cancel(molCu))xx(107.9Ag)/(1cancel(molAg))=68.0gAg