If 9.85 grams of copper metal react with 31.0 grams of silver nitrate, how many grams of copper nitrate can be formed and how many grams of the excess reactant will be left over when the reaction is complete? show all of your work. unbalanced equation: cu + agno3 yields cu(no3)2 + ag

Cu + 2AgNO3 = Cu(NO3)2 + 2Ag

9.85 g Cu ( 1 mol Cu / 63.55 g Cu ) = 0.15 mol Cu
31.0 g AgNO3 ( 1 mol AgNO3 / 169.87 g AgNO3 ) = 0.18 mol AgNO3

limiting reactant (AgNO3)

0.18 mol AgNO3 ( 1 mol Cu(NO3)2 / 2 mol AgNO3 ) (187.56 g / 1 mol) =16.88 g Cu(NO3)2

0.15 mol Cu - 0.18 mol AgNO3 ( 1 mol Cu / 2 mol AgNo3) = 0.06 mol Cu

0.06 mol Cu ( 63.55 g Cu / 1 mol Cu ) = 3.81 g Cu

The balanced chemical reaction: 

Cu+2AgNO3=Cu(NO3)2+2Ag

We are given the amount of the reactants to be used for the reaction. These values will be the starting point of our calculations.

9.85 g Cu ( 1 mol Cu / 63.55 g Cu ) = 0.15 mol Cu
31.0 g AgNO3 (1 mol AgNO3 / 169.87 g AgNO3 ) = 0.18 mol AgNO3

The limiting reactant is AgNO3.

0.18 mol AgNO3 (1 mol Cu(NO3)2 / 2 mol AgNO3 ) (187.56 g / 1 mol ) =16.88 g Cu(NO3)2

0.15 mol Cu - 0.18 mol AgNO3 ( 1 mol Cu / 2 mol AgNO3 ) = 0.06 mol Cu excess

0.06 mol Cu ( 63.55 g Cu / 1 mol Cu ) = 3.81 g Cu excess

Hope this helps! :)

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