Question

How many moles of Ag can be produced if 350. g of Cu are reacted with excess AgNO3 according to the equation Cu(s) + 2AgNO3(aq) ® 2Ag(s) + Cu(NO3)2(aq)?

Answer 1

Answer: The moles of Ag produced will be 11.02 moles.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

For copper:

Given mass of copper = 350 g

Molar mass of copper = 63.5 g/mol

Putting values in above equation, we get:

$\text{Moles of copper}=\frac{350g}{63.5g/mol}=5.51mol$

For the given chemical equation:

$Cu(s)+2AgNO_3(aq.)\rightarrow 2Ag(s)+Cu(NO_3)_2(aq.)$

Silver nitrate is present in excess, so it is considered as an excess reagent and copper is considered as a limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

1 mole of copper produces 2 moles of silver.

So, 5.51 moles of copper will produce = $\frac{2}{1}\times 5.51=11.02mol$ of silver.

Thus, the moles of Ag produced will be 11.02 moles.

Answer 2

Hello.

balance the equation first 
Cu + 2AgNO3 ----> [2Ag] + Cu(NO3)2 


Now set up the equation 
350gCu*(1molCu/ 63.546g Cu) * (2molAg/1molCu) * (107.8682g Ag/1molAg)
 

Cancel out 1molCu with 1molCu, and cancel out 2molAg with 1molAg (it still remains 2mol Ag) 

You are left with 
350gCu*(1/63.546g Cu)*(2Ag/1)* (107.8682g Ag/1) 


350gCu*1*2*107.8682= 75507.74g 


now divide 75507.74g by 63.546g 

you get [1188.2374972460896g Ag]

Have a nice day