Question

If 6.51 g of copper is reacted with 28.4 g of silver nitrate, the products will be copper (ii) nitrate and silver metal. what is the theoretical mass of silver that will be produced?

Answer 1

`m_(Ag)=18.0gAg`

Step-by-step explanation:

Hello,

In this case, the undergoing chemical reaction is:

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

Now, since the theoretical yield of silver is required, we first must identify the limiting reactant by comparing the available moles of copper and the moles of copper that are consumed by the 28.4 g of silver nitrate as shown below:

`n_(Cu)^(available)=6.51gCu*(1molCu)/(63.55gCu) =0.102molCu\\n_(Cu)^(consumed)=28.4gAgNO_3*(1molAgNO_3)/(170gAgNO_3)*(1molCu)/(2molAgNO_3)=0.0835molCu`

Hence, since there are more available moles of copper than consumed, we understand copper as the excess reactant and silver nitrate as the limiting one, therefore, the theoretical yield of silver turns out:

`m_(Ag)=0.0835molCu*(2molAg)/(1molCu)*(108gAg)/(1molAg) \\m_(Ag)=18.0gAg`

Best regards.

Answer 2

The theoretical mass of silver that will be produced is calculated as follows

write the reaction equation
Cu + 2Ag(No3) ---> 2Ag + Cu(NO3)2

calculated the moles of each reagent
moles =mass/molar mass

Cu = 6.51g/63.5g/mol= 0.103 moles
Ag(NO3)2 = 28.4 g/169.87g/mol = 0.167 moles

Cu is the limiting reagent therefore the moles Ag= 2 x0.167 = 0.334 moles

mass = moles xmolar mass
0.334mol x 107.87g/mol = 36.02 grams