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

Answer:

0.252 mol

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:

$16.0gCu.\frac{1molCu}{63.55gCu} .\frac{1molAg}{1molCu} =0.252 molAg$

Answer 2

Answer:

$n_{Ag} =0.504molAg$

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:

$n_{Ag} =16.0gCu*\frac{1molCu}{63.5gCu}*\frac{2molAg}{1molCu}\\n_{Ag} =0.504molAg$

Best regards.