3 moles of Silver is formed
       1 mole of excess AgNO₃ is left unreacted
The balance chemical equation is as follow,
            Cu  +  2 AgNO₃   →   Cu(NO₃)₂  +  2 Ag
Step 1: Finding out Limiting Reagent:
According to equation,
         1 mole of Cu reacts with  =  2 moles of AgNO₃
So,
       1.5 moles of Cu will react with  =  X moles of AgNO₃
Solving for X,
          X  =  (1.5 mol × 2 mol) ÷ 1 mol
          X  =  3 mol of AgNO₃
Therefore, 1.5 moles of Copper requires 3 moles of AgNO₃ for complete reaction but, we are provided with 4 moles of AgNO₃ which is in excess hence, Cu is the limiting reagent and will control the yield of products.
Step 2: Calculate Amount of Ag formed:
According to equation,
           1 mole of Cu produced  =  2 moles of Ag
So,
         1.5 moles of Cu will produce  =  X moles of Ag
Solving for X,
           X  =  (1.5 mol × 2 mol) ÷ 1 mol
           X  =  3 moles of Silver
Step 3: Calculating Excess AgNO₃ left:
As the reaction required only 3 moles of AgNO₃ and we were provided with 4 moles of AgNO₃ so the amount left was,
              Excess AgNO₃  =  4 moles - 3 moles
              Excess AgNO₃  = 1 mole
Answer:
1 mole of excess reactant (AgNO[tex]_3[/tex] will be left.
Step-by-step explanation:
First of all, we need to balance the given equation to get:
[tex]Cu + [/tex]2[tex]AgNO^3[/tex] → [tex]Cu(NO^3)^2+[/tex]2[tex]Ag[/tex]
We know that,
1 mole of Cu reacts with 2 moles of AgNO[tex]_3[/tex]
So supposing the number of moles of AgNO[tex]_3[/tex] to be x for 1.5 moles of Cu, we can find x:
[tex]x = \frac{1.5*2}{1} = 3[/tex]
Therefore 3 moles of AgNO[tex]_3[/tex] are needed to complete the reaction while we have excess of it (4 moles). So 1 mole of excess reactant AgNO[tex]_3[/tex] will be left.
