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?

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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?

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Answer 1 · 2015-12-04T18:42:11

#68.0g# of silver.

Explanation:

The reaction is:

#Cu(s)+AgNO_3(aq)->Ag(s)+CuNO_3(aq)#

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#

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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?

1 Answer

Explanation:

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