Question

If 1.5 moles of copper metal react with 4.0 moles of silver nitrate, how many moles of silver metal can be formed, and how many moles of the excess reactant will be left over when the reaction is complete?
Unbalanced equation: Cu + AgNO3 → Cu(NO3)2 + Ag .

Answer 1

Answer:

1 mole of excess reactant (AgNO$_3$ will be left.

Step-by-step explanation:

First of all, we need to balance the given equation to get:

$Cu +$2$AgNO^3$ → $Cu(NO^3)^2+$2$Ag$

We know that,

1 mole of Cu reacts with 2 moles of AgNO$_3$

So supposing the number of moles of AgNO$_3$ to be x for 1.5 moles of Cu, we can find x:

$x = \frac{1.5*2}{1} = 3$

Therefore 3 moles of AgNO$_3$ are needed to complete the reaction while we have excess of it (4 moles). So 1 mole of excess reactant AgNO$_3$ will be left.

Answer 2

Answers:

              3 moles of Silver is formed

              1 mole of excess AgNO₃ is left unreacted

Solution:

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