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

Answer:

$m_{Ag}=18.0gAg$

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:

$n_{Cu}^{available}=6.51gCu*\frac{1molCu}{63.55gCu} =0.102molCu\\n_{Cu}^{consumed}=28.4gAgNO_3*\frac{1molAgNO_3}{170gAgNO_3}*\frac{1molCu}{2molAgNO_3}=0.0835molCu$

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:

$m_{Ag}=0.0835molCu*\frac{2molAg}{1molCu}*\frac{108gAg}{1molAg} \\m_{Ag}=18.0gAg$

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