how many grams of each species will be present at the end when 0.500g of copper is added to 2.500g of AgNO3?
Cu(s) + 2 AgNO3(aq) --> 2 Ag(s) + Cu(NO3)2 (aq)
Here is a worked example of a stoichiometry problem.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To find the number of grams of each species at the end of the reaction, we need to determine the limiting reactant first. The limiting reactant is the reactant that is completely consumed, which will determine the amount of product formed.
1. Calculate the number of moles for each reactant:
- Moles of copper (Cu):
Cu = (mass of Cu) / (molar mass of Cu)
Cu = 0.500 g / 63.55 g/mol = 0.00787 mol
- Moles of silver nitrate (AgNO3):
AgNO3 = (mass of AgNO3) / (molar mass of AgNO3)
AgNO3 = 2.500 g / (107.87 g/mol + 14.01 g/mol + 3 * 16.00 g/mol) = 0.0176 mol
2. Determine the mole ratio between Cu and AgNO3 from the balanced chemical equation:
Cu:AgNO3 = 1:2
3. Calculate the number of moles of AgNO3 needed to react with Cu:
Moles of AgNO3 needed = 2 * Moles of Cu = 2 * 0.00787 mol = 0.0157 mol
4. Compare the number of moles of AgNO3 needed with the actual moles of AgNO3 available:
Since 0.0157 mol of AgNO3 is needed, which is greater than the 0.0176 mol available, AgNO3 is in excess and Cu is the limiting reactant.
5. Calculate the number of moles of Ag and Cu(NO3)2 formed from the reaction:
Moles of Ag = 2 * Moles of Cu = 2 * 0.00787 mol = 0.0157 mol
Moles of Cu(NO3)2 = Moles of Cu = 0.00787 mol
6. Calculate the mass of each species formed:
Mass of Ag = Moles of Ag * (molar mass of Ag)
Mass of Ag = 0.0157 mol * 107.87 g/mol = 1.69 g
Mass of Cu(NO3)2 = Moles of Cu(NO3)2 * (molar mass of Cu(NO3)2)
Mass of Cu(NO3)2 = 0.00787 mol * (63.55 g/mol + 2 * 14.01 g/mol + 6 * 16.00 g/mol) = 1.38 g
To determine how many grams of each species will be present at the end of the reaction, we need to use stoichiometry. Stoichiometry is a branch of chemistry that deals with the quantitative relationship between reactants and products in a chemical reaction.
In this reaction, the stoichiometric ratio between copper (Cu) and silver nitrate (AgNO3) is 1:2, which means that one mole of copper reacts with two moles of silver nitrate.
Step 1: Convert the mass of copper to moles
To do this, we need to know the molar mass of copper, which is 63.55 g/mol.
Using the formula: moles = mass / molar mass, we can calculate the moles of copper:
moles of Cu = 0.500 g / 63.55 g/mol
Step 2: Determine the limiting reagent
To determine the limiting reagent, we compare the moles of copper to the moles of silver nitrate. Since the stoichiometric ratio between them is 1:2, we need to multiply the moles of copper by 2 to get the moles of silver nitrate required to react completely:
moles of AgNO3 required = 2 * moles of Cu
Now, compare the moles of AgNO3 required to the moles of AgNO3 given (2.500 g).
moles of AgNO3 given = 2.500 g / molar mass of AgNO3
If the moles of AgNO3 given are greater than the moles of AgNO3 required, AgNO3 is in excess, and copper will be the limiting reagent. If the moles of AgNO3 required are greater, then AgNO3 is the limiting reagent.
Step 3: Calculate the moles of each species present at the end
a) If copper is the limiting reagent:
Since copper is consumed in the reaction, the moles of copper at the end will be zero. The moles of silver formed can be calculated using the stoichiometric ratio:
moles of Ag = 2 * moles of Cu
b) If silver nitrate is the limiting reagent:
The moles of copper will be consumed until all of the silver nitrate is used up. The moles of copper nitrate formed can be calculated using the stoichiometric ratio:
moles of Cu(NO3)2 = moles of AgNO3 given - (2 * moles of Cu)
To convert the moles to grams, multiply the moles by the molar mass of each species.
By following these steps, you can determine the grams of each species present at the end of the reaction.