Question

Given the following reaction: \ce{Cu + 2AgNO3 -> 2Ag + Cu(NO3)2}Cu+2AgNO3 ​ ​ 2Ag+Cu(NOX 3 ​ )X 2 ​ How many moles of \ce{Ag}AgA, g will be produced from 16.0 \text{ g}16.0 g16, point, 0, start text, space, g, end text of \ce{Cu}CuC, u, assuming \ce{AgNO3}AgNO3 ​ is available in excess

Answer 1

0.252 mol

Step-by-step explanation:

Given the following reaction:

Cu + 2 AgNO₃ → 2 Ag + Cu(NO₃)₂

How many moles of Ag will be produced from 16.0 g Cu, assuming AgNO3 ​ is available in excess.

First, we write the balanced equation.

Cu + 2 AgNO₃ → 2 Ag + Cu(NO₃)₂

We can establish the following relations.

• The molar mass of Cu is 63.55 g/mol.
• The molar ratio of Cu to Ag is 1:1.

The moles of Ag produced from 16.0 g of Cu are:

`16.0gCu.(1molCu)/(63.55gCu) .(1molAg)/(1molCu) =0.252 molAg`

Answer 2

`n_(Ag) =0.504molAg`

Step-by-step explanation:

Hello,

In this case, the undergoing chemical reaction is:

`Cu+2AgNO_3-->Cu(NO_3)_2+2Ag`

As the silver nitrate is in excess, the yielded moles of silver are computed from the 16g of copper as shown below, including the corresponding stoichiometric 1 to 2 relationship between copper and silver respectively:

`n_(Ag) =16.0gCu*(1molCu)/(63.5gCu)*(2molAg)/(1molCu)\\n_(Ag) =0.504molAg`

Best regards.