Question

Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A solution containing \(3.50 \mathrm{~g}\) of sodium carbonate is mixed with one containing \(5.00 \mathrm{~g}\) of silver nitrate. How many grams of sodium carbonate, silver nitrate, silver carbonate, and sodium nitrate are present after the reaction is complete?

Short answer

After the reaction is complete, there are 1.56 g of sodium carbonate, 0 g of silver nitrate, 4.05 g of silver carbonate, and 2.50 g of sodium nitrate.

Step-by-step solution

1. Write the balanced chemical equation for the reaction

The balanced chemical equation for the reaction between sodium carbonate (Na2CO3) and silver nitrate (AgNO3) is: \( Na_2CO_3(aq) + 2AgNO_3(aq) \rightarrow 2NaNO_3(aq) + Ag_2CO_3(s) \)

2. Calculate the molar masses of the compounds involved in the reaction

We need to find the molar masses of sodium carbonate (Na2CO3), silver nitrate (AgNO3), silver carbonate (Ag2CO3), and sodium nitrate (NaNO3). Molar mass of Na2CO3: (Na: 22.99 g/mol, C: 12.01 g/mol, O: 16.00 g/mol) \(1 \cdot 2 \cdot 22.99 + 1 \cdot 12.01 + 3 \cdot 16.00 = 105.99 \mathrm{g/mol}\) Molar mass of AgNO3: (Ag: 107.87 g/mol, N: 14.01 g/mol, O: 16.00 g/mol) \(1 \cdot 107.87 + 1 \cdot 14.01 + 3 \cdot 16.00 = 169.87 \mathrm{g/mol}\) Molar mass of Ag2CO3: (Ag: 107.87 g/mol, C: 12.01 g/mol, O: 16.00 g/mol) \(2 \cdot 107.87 + 1 \cdot 12.01 + 3 \cdot 16.00 = 275.75 \mathrm{g/mol}\) Molar mass of NaNO3: (Na: 22.99 g/mol, N: 14.01 g/mol, O: 16.00 g/mol) \(1 \cdot 22.99 + 1 \cdot 14.01 + 3 \cdot 16.00 = 85.00 \mathrm{g/mol}\)

3. Determine the initial moles of sodium carbonate and silver nitrate

Now that we have the molar masses, we can determine the moles of sodium carbonate and silver nitrate before the reaction. Moles of Na2CO3: \( \frac{3.50 \mathrm{~g}}{105.99 \frac{\mathrm{g}}{\mathrm{mol}}} = 0.0330 \mathrm{~mol} \) Moles of AgNO3: \( \frac{5.00 \mathrm{~g}}{169.87 \frac{\mathrm{g}}{\mathrm{mol}}} = 0.0294 \mathrm{~mol} \)

4. Identify the limiting reactant and calculate the moles of each product

To find the limiting reactant, we'll compare the mole ratio of sodium carbonate and silver nitrate with the coefficients in the balanced equation. The limiting reactant will be the one that produces the least amount of product. Mole ratio: \( \frac{0.0330 \mathrm{~mol \, Na_2CO_3}}{0.0294 \mathrm{~mol \, AgNO_3}} = 1.12\) Since the mole ratio is greater than 1, and the balanced equation shows a 1:2 ratio, silver nitrate (AgNO3) is the limiting reactant. Now, we can calculate the moles of sodium nitrate and silver carbonate produced using the coefficients and moles of the limiting reactant. Moles of NaNO3: \( 0.0294 \mathrm{~mol \, AgNO_3} \cdot \frac{2\, \mathrm{mol \, NaNO_3}}{2\, \mathrm{mol \, AgNO_3}} = 0.0294 \mathrm{~mol \, NaNO_3} \) Moles of Ag2CO3: \( 0.0294 \mathrm{~mol \, AgNO_3} \cdot \frac{1\, \mathrm{mol \, Ag_2CO_3}}{2\, \mathrm{mol \, AgNO_3}} = 0.0147 \mathrm{~mol \, Ag_2CO_3} \)

5. Calculate the grams of each compound after the reaction

Now, we can calculate the grams of each compound after the reaction is complete. Mass of unreacted Na2CO3: Since we used up all the limiting reactant (silver nitrate), there will be some unreacted sodium carbonate left. To find this mass, we can use the stoichiometry of the balanced equation and the moles of silver nitrate that reacted. \( 0.0294 \mathrm{~mol \, AgNO_3} \cdot \frac{1\, \mathrm{mol \, Na_2CO_3}}{2\, \mathrm{mol \, AgNO_3}} = 0.0147 \mathrm{~mol \, Na_2CO_3} \) \( 0.0147 \mathrm{~mol \, Na_2CO_3} \cdot 105.99 \frac{\mathrm{g}}{\mathrm{mol}} = 1.56 \mathrm{~g \, Na_2CO_3} \) Mass of Ag2CO3 formed: \( 0.0147 \mathrm{~mol \, Ag_2CO_3} \cdot 275.75 \frac{\mathrm{g}}{\mathrm{mol}} = 4.05 \mathrm{~g \, Ag_2CO_3} \) Mass of NaNO3 formed: \( 0.0294 \mathrm{~mol \, NaNO_3} \cdot 85.00 \frac{\mathrm{g}}{\mathrm{mol}} = 2.50\mathrm{~g \, NaNO_3} \) Since all silver nitrate reacted, there's 0 g of silver nitrate left after the reaction.

6. Summary of the final amounts of each compound

After the reaction is complete, the amounts of each compound are as follows: - Sodium carbonate (Na2CO3): 1.56 g - Silver nitrate (AgNO3): 0 g - Silver carbonate (Ag2CO3): 4.05 g - Sodium nitrate (NaNO3): 2.50 g

Key concepts

balanced chemical equation

A balanced chemical equation is a fundamental tool in understanding chemical reactions. It provides vital information about the reactants and products, along with their stoichiometric relationships. In the context of our reaction, we have sodium carbonate (\( \text{Na}_2\text{CO}_3 \)) and silver nitrate (\( \text{AgNO}_3 \)) reacting to form silver carbonate (\( \text{Ag}_2\text{CO}_3 \)) and sodium nitrate (\( \text{NaNO}_3 \)).

The balanced equation for this reaction is:\[ \text{Na}_2\text{CO}_3 (aq) + 2\text{AgNO}_3 (aq) \rightarrow 2\text{NaNO}_3 (aq) + \text{Ag}_2\text{CO}_3 (s) \]

• The balanced equation signifies that 1 mole of sodium carbonate reacts with 2 moles of silver nitrate.
• This results in 2 moles of sodium nitrate and 1 mole of silver carbonate.

It is crucial to ensure that the number of atoms of each element is the same on both sides of the equation, fulfilling the law of conservation of mass. Balancing the equation helps us accurately predict the amounts of products formed from given quantities of reactants.

stoichiometry

Stoichiometry plays a key role in predicting the quantitative aspects of chemical reactions. It uses the balanced chemical equation to relate the amounts of reactants and products.

In our problem, stoichiometry allows us to:

• Identify the limiting reactant by comparing the mole ratios of sodium carbonate and silver nitrate from the balanced equation. The limiting reactant is the substance that completely reacts and limits the amount of product formed.
• Calculate the expected moles of products formed.

For the reaction at hand:

• Sodium carbonate's initial amount is 0.0330 mol.
• Silver nitrate's initial amount is 0.0294 mol.

Since the stoichiometric ratio in the balanced equation requires 1 mole of sodium carbonate to react with 2 moles of silver nitrate, silver nitrate is found to be the limiting reactant. This influences how much silver carbonate and sodium nitrate can be produced.

molar mass calculation

Understanding molar masses of the involved substances is pivotal for stoichiometry. These values convert the mass of each substance into moles, which is used in calculations.

For the relevant compounds:

• **Sodium carbonate (\( \text{Na}_2\text{CO}_3 \))**: 105.99 g/mol
• **Silver nitrate (\( \text{AgNO}_3 \))**: 169.87 g/mol
• **Silver carbonate (\( \text{Ag}_2\text{CO}_3 \))**: 275.75 g/mol
• **Sodium nitrate (\( \text{NaNO}_3 \))**: 85.00 g/mol

These molar masses are calculated by summing the weighted average atomic masses of each element (from the periodic table) times the number of each atom in the molecule.

For example, the molar mass of sodium carbonate is calculated by:\[ 2 \times 22.99 + 12.01 + 3 \times 16.00 = 105.99 \, \text{g/mol} \]Knowledge of molar masses lets us make meaningful comparisons and convert masses to moles for further stoichiometric evaluations.

chemical reactions in solutions

Chemical reactions in solutions are unique because reactants are often dissolved in a liquid medium, usually water, which allows ions to move freely and react.

For our exercise:

• Sodium carbonate and silver nitrate are dissolved in water, making them easily available for reaction due to their dissociation into ions.
• This environment promotes the formation of insoluble silver carbonate, which precipitates out as a solid, in a classic precipitation reaction.
• The soluble sodium nitrate remains dissolved in the aqueous solution, as it is highly soluble in water.

Reaction dynamics in solutions are chiefly influenced by the solubility rules and the behavior of the dissolved ions. Understanding how solubilities affect the reaction's outcome is essential for accurately predicting and calculating the amounts of substances before and after the reaction.