Question

Balance the following reaction.

Cu + Ag(NO3) --> Cu(NO3)2 + Ag

If you are given 2.5 grams of copper and 5 g of silver nitrate


What is the Limiting Reactant? __________ (Cu/AgNO3)

How much of each product do you produce?

_________ g Cu(NO3)2

_________ g Ag

Answer 1

The balanced chemical equation for the reaction is:

2 Cu + Ag(NO3)2 -> 2 Cu(NO3)2 + Ag

First, let's convert the amount of reactants into moles:

2.5 g of Cu = 2.5 g / 63.55 g/mol = 0.0395 moles of Cu
5 g of AgNO3 = 5 g / 169.87 g/mol = 0.0294 moles of AgNO3

The limiting reactant is AgNO3, as it is the reactant that is present in the smallest amount.

Now, we can use the stoichiometry of the reaction to find the amount of each product produced:

Ag: 0.0294 moles * 1 mole/2 moles = 0.0147 moles = 0.0147 * 107.87 g/mol = 1.58 g
Cu(NO3)2: 0.0294 moles * 2 moles/2 moles = 0.0294 moles = 0.0294 * 164.48 g/mol = 4.87 g

So, you will produce 1.58 g of Ag and 4.87 g of Cu(NO3)2.

Answer 2

The balanced chemical equation for this reaction is:

2 Cu + Ag(NO3)2 -> 2 Cu(NO3)2 + Ag

The limiting reactant is the reactant that runs out first and determines the amount of product that can be produced. To determine the limiting reactant, we need to calculate the number of moles of each reactant.

From the given mass of copper, we can calculate the number of moles as follows:

2.5 g Cu / 63.55 g/mol = 0.0395 mol Cu

From the given mass of silver nitrate, we can calculate the number of moles as follows:

5 g AgNO3 / 169.87 g/mol = 0.0295 mol AgNO3

Since 0.0295 mol of AgNO3 is less than 0.0395 mol of Cu, AgNO3 is the limiting reactant.

To calculate the amount of each product produced, we can use the stoichiometry of the balanced equation.

For Ag, the reaction produces 1 mole of Ag for every 2 moles of AgNO3, so we have:

0.0295 mol AgNO3 * 1 mole Ag / 2 moles AgNO3 = 0.0148 mol Ag

And converting moles to grams:

0.0148 mol Ag * 107.87 g/mol = 1.59 g Ag

For Cu(NO3)2, the reaction produces 2 moles of Cu(NO3)2 for every 2 moles of AgNO3, so we have:

0.0295 mol AgNO3 * 2 moles Cu(NO3)2 / 2 moles AgNO3 = 0.0295 mol Cu(NO3)2

And converting moles to grams:

0.0295 mol Cu(NO3)2 * 164.46 g/mol = 4.86 g Cu(NO3)2

Therefore, the limiting reactant is AgNO3 (Cu/AgNO3) and the reaction produces 1.59 g of Ag and 4.86 g of Cu(NO3)2.