If 2.50 mol of copper and 5.50 mol of silver nitrate are available to react by a single replacement:Cu + 2AgNO3 -> Cu(NO3)2 + 2Ag
What is the amount in moles of each product formed?
Anytime BOTH of the reactants are listed the problem is a limiting reagent one.
Convert moles Cu to moles of the products.
Convert 5.5 moles AgNO3 to moles of the products.
The reagent producing the smaller number of moles of the products is the limiting reagent and the smaller number of moles will be the moles of the products after the reaction is complete.
To find the amount in moles of each product formed, we can use the stoichiometry of the balanced chemical equation.
The balanced equation is:
Cu + 2AgNO3 -> Cu(NO3)2 + 2Ag
From the equation, we can see that the mole ratio between copper (Cu) and copper nitrate (Cu(NO3)2) is 1:1, and the mole ratio between silver nitrate (AgNO3) and silver (Ag) is 2:2 or simplified to 1:1.
Given that 2.50 mol of copper is available, this means that 2.50 mol of copper nitrate will be formed.
Similarly, since the mole ratio is 1:1 between silver nitrate and silver, the amount of silver formed will also be 2.50 mol.
Therefore, the amount in moles of each product formed would be:
Cu(NO3)2: 2.50 mol
Ag: 2.50 mol
To determine the amount in moles of each product formed, we can use the stoichiometry of the reaction. The balanced equation tells us that:
1 mole of Cu reacts with 2 moles of AgNO3 to produce 1 mole of Cu(NO3)2 and 2 moles of Ag.
Therefore, we need to calculate the amount of each product formed based on the given amounts of reactants.
Given:
- 2.50 mol of Cu
- 5.50 mol of AgNO3
Let's start with calculating the amount of Cu(NO3)2 formed:
From the balanced equation, we see that 1 mole of Cu reacts to produce 1 mole of Cu(NO3)2.
So, the amount of Cu(NO3)2 formed will be equal to the amount of Cu used in the reaction, which is 2.50 mol.
Now, let's calculate the amount of Ag formed:
From the balanced equation, we see that 1 mole of Cu reacts to produce 2 moles of Ag.
Therefore, the amount of Ag formed will be twice the amount of Cu used in the reaction, which is 2.50 mol.
So, the amount of each product formed is:
- 2.50 mol of Cu(NO3)2
- 5.00 mol of Ag