Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A solution containing 3.30 g of sodium carbonate is mixed with one containing 3.00 g of silver nitrate. How many grams of each of the following compounds are present after the reaction is complete?
sodium carbonate?
silver nitrate?
silver carbonate?
sodium nitrate?
Na2CO3 + 2AgNO3 ==> 2NaNO3 + Ag2CO3
This is a limiting reagent problem. You know that because amounts are given for BOTH reactants. First you must determine the limiting reagent.
mols Na2CO3 = grams/molar mass.
mols AgNO3 = grams/molar mass.
Using the coefficients in the balanced equation, convert mols Na2CO3 to mols NaNO3.
Do the same to convert mols AgNO3 to mols NaNO3.
It is likely that the two numbers will not be the same; obviously one of them must be wrong. The correct value in limiting reagent problems is ALWAYS the smaller one and the reagent providing that value is the limiting reagent.
Now that you know the limiting reagent, use mols of that reagent and convert to mols Ag2CO3. Convert mols Ag2CO3 and mols NaNO3 to grams by g = mols x molar mass.
The limiting reagent will be zero after the rxn is over.
To determine the amount of the non-limiting reagent after the rxn, convert mols of the limiting reagent to mols of the non-limiting reagent, convert that to grams, subtract from initial grams to arrive at the amount remaining unreacted. Post your work if you get stuck.
Could you show me step-by-step how to find the limiting reagent? That's the part I get a little confused in.
To determine the grams of each compound present after the reaction is complete, we need to use stoichiometry and the balanced equation for the reaction:
Na2CO3 + 2AgNO3 -> Ag2CO3 + 2NaNO3
To find the limiting reagent, we compare the number of moles of sodium carbonate and silver nitrate.
1. Calculate the number of moles for sodium carbonate:
Molar mass of Na2CO3 = 22.99 g/mol + 12.01 g/mol + (3 * 16.00 g/mol) = 105.99 g/mol
Moles of Na2CO3 = 3.30 g / 105.99 g/mol = 0.0311 mol
2. Calculate the number of moles for silver nitrate:
Molar mass of AgNO3 = 107.87 g/mol + 14.01 g/mol + (3 * 16.00 g/mol) = 169.87 g/mol
Moles of AgNO3 = 3.00 g / 169.87 g/mol = 0.0177 mol
3. Determine the limiting reagent:
The ratio of Na2CO3 to AgNO3 is 1:2 in the balanced equation.
Since the ratio of moles of Na2CO3 to AgNO3 is approximately 1.75:1, Na2CO3 is the limiting reagent.
4. Calculate the products formed:
a. Moles of Ag2CO3 = Moles of limiting reagent (Na2CO3) = 0.0311 mol
b. Moles of NaNO3 = 2 * Moles of Ag2CO3 = 2 * 0.0311 mol = 0.0622 mol
5. Calculate the grams of each compound:
a. Mass of Na2CO3 = Moles of Na2CO3 * Molar mass of Na2CO3
Mass of Na2CO3 = 0.0311 mol * 105.99 g/mol = 3.28 g
b. Mass of AgNO3 = Moles of AgNO3 * Molar mass of AgNO3
Mass of AgNO3 = 0.0177 mol * 169.87 g/mol = 3.01 g
c. Mass of Ag2CO3 = Moles of Ag2CO3 * Molar mass of Ag2CO3
Mass of Ag2CO3 = 0.0311 mol * 275.75 g/mol = 8.56 g
d. Mass of NaNO3 = Moles of NaNO3 * Molar mass of NaNO3
Mass of NaNO3 = 0.0622 mol * 85.0 g/mol = 5.29 g
So, after the reaction is complete, the grams of each compound present are:
- Sodium carbonate: 3.28 g
- Silver nitrate: 3.01 g
- Silver carbonate: 8.56 g
- Sodium nitrate: 5.29 g
To find out how many grams of each compound are present after the reaction is complete, we need to determine the limiting reactant first. The limiting reactant is the reactant that is completely consumed and determines the amount of product formed.
1. Starting with the given amounts of sodium carbonate and silver nitrate:
- Sodium carbonate (Na2CO3): 3.30 g
- Silver nitrate (AgNO3): 3.00 g
2. Calculate the number of moles for each reactant:
- Moles of sodium carbonate = mass / molar mass
- Moles of sodium carbonate = 3.30 g / (2(22.99 g/mol) + 12.01 g/mol + 3(16.00 g/mol)) = 0.0392 mol
- Moles of silver nitrate = 3.00 g / (107.87 g/mol + 14.01 g/mol + 3(16.00 g/mol)) = 0.0146 mol
3. Set up a balanced chemical equation for the reaction:
- Na2CO3 + 2AgNO3 -> Ag2CO3 + 2NaNO3
4. Use the coefficients of the balanced equation to determine the stoichiometry:
- From the balanced equation, we can see that 1 mole of sodium carbonate reacts with 2 moles of silver nitrate to form 1 mole of silver carbonate and 2 moles of sodium nitrate.
5. Determine the limiting reactant:
- The stoichiometry ratio tells us that 1 mole of sodium carbonate reacts with 2 moles of silver nitrate. Since we have fewer moles of silver nitrate (0.0146 mol) compared to sodium carbonate (0.0392 mol), silver nitrate is the limiting reactant.
6. Calculate the amount of each compound present after the reaction:
- Since silver nitrate is the limiting reactant, it will be completely consumed in the reaction. Therefore, we can calculate the amount of sodium carbonate, silver carbonate, and sodium nitrate formed based on the stoichiometry of the balanced equation.
- Moles of sodium carbonate remaining = Moles of sodium carbonate initially - Moles of sodium carbonate reacted
- Moles of sodium carbonate remaining = 0.0392 mol - (1 mol of Na2CO3 reacts with 2 mol of AgNO3, so 0.0146 mol of Na2CO3 reacts) = 0.0246 mol
- Moles of silver carbonate formed = Moles of silver nitrate reacted (according to the stoichiometry ratio)
- Moles of silver carbonate formed = 0.0146 mol
- Moles of sodium nitrate formed = Moles of silver nitrate reacted * 2 (according to the stoichiometry ratio)
- Moles of sodium nitrate formed = 0.0146 mol * 2 = 0.0292 mol
7. Calculate the mass of each compound formed:
- Mass = moles * molar mass
- Mass of sodium carbonate remaining = 0.0246 mol * (2(22.99 g/mol) + 12.01 g/mol + 3(16.00 g/mol)) = 2.16 g
- Mass of silver carbonate formed = 0.0146 mol * (107.87 g/mol + 14.01 g/mol + 3(16.00 g/mol)) = 2.03 g
- Mass of sodium nitrate formed = 0.0292 mol * (22.99 g/mol + 14.01 g/mol + 3(16.00 g/mol)) = 1.99 g
8. Finally, the masses of each compound after the reaction is complete are:
- Sodium carbonate: 2.16 g remaining
- Silver nitrate: completely consumed
- Silver carbonate: 2.03 g formed
- Sodium nitrate: 1.99 g formed