Problem 81 - Chemistry The Central Science In Si Units

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?

Step-by-Step Solution

Answer

1.94 g sodium carbonate, 0 g silver nitrate, 4.05 g silver carbonate, 2.50 g sodium nitrate remain.

Step 1: Write the Balanced Equation

To determine the products of the reaction, write the balanced chemical equation: \[ \text{Na}_2\text{CO}_3 (aq) + 2\text{AgNO}_3 (aq) \rightarrow \text{Ag}_2\text{CO}_3 (s) + 2\text{NaNO}_3 (aq) \] This shows 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.

Step 2: Calculate the Molar Masses

Find the molar mass of each compound: - Sodium carbonate (Na₂CO₃): \(2(22.99) + 12.01 + 3(16.00) = 105.99\text{ g/mol}\)- Silver nitrate (AgNO₃): \(107.87 + 14.01 + 3(16.00) = 169.87\text{ g/mol}\)- Silver carbonate (Ag₂CO₃): \(2(107.87) + 12.01 + 3(16.00) = 275.75\text{ g/mol}\)- Sodium nitrate (NaNO₃): \(22.99 + 14.01 + 3(16.00) = 84.99\text{ g/mol}\)

Step 3: Convert Gram to Moles

Convert mass of reactants to moles:- Moles of sodium carbonate: \( \frac{3.50\text{ g}}{105.99\text{ g/mol}} \approx 0.0330\text{ mol} \)- Moles of silver nitrate: \( \frac{5.00\text{ g}}{169.87\text{ g/mol}} \approx 0.0294\text{ mol} \)

Step 4: Determine the Limiting Reactant

Based on the stoichiometry (1:2 ratio), sodium carbonate would require twice the moles of silver nitrate to react completely. We have:- Sodium carbonate requires: \(0.0330 \times 2 = 0.0660\text{ mol} \) of silver nitrate.- Available silver nitrate: \(0.0294\text{ mol} \), which is less than required.Thus, silver nitrate is the limiting reactant.

Step 5: Calculate the Products form Limiting Reactant

Using stoichiometry, calculate moles of products formed using the limiting reactant (Silver Nitrate):- Moles of silver carbonate formed (1:2 ratio with silver nitrate): \(0.0294\text{ mol AgNO}_3 \rightarrow \frac{0.0294}{2} = 0.0147\text{ mol Ag}_2\text{CO}_3 \)- Moles of sodium nitrate formed (2:2 ratio with silver nitrate): \(0.0294\text{ mol} \text{AgNO}_3 \rightarrow 0.0294\text{ mol NaNO}_3 \)

Step 6: Convert Moles to Grams of Products

Convert moles of products to grams:- Mass of silver carbonate: \(0.0147\text{ mol} \times 275.75\text{ g/mol} \approx 4.05\text{ g} \)- Mass of sodium nitrate: \(0.0294\text{ mol} \times 84.99\text{ g/mol} \approx 2.50\text{ g} \)

Step 7: Calculate Excess Reactant Remaining

Calculate moles and mass of sodium carbonate left unreacted:- Sodium carbonate initially: \(0.0330\text{ mol} \)- Sodium carbonate used: \( \frac{0.0294}{2} = 0.0147\text{ mol} \)- Sodium carbonate remaining: \(0.0330 - 0.0147 = 0.0183\text{ mol} \)- Mass remaining: \(0.0183\text{ mol} \times 105.99\text{ g/mol} \approx 1.94\text{ g} \)

Key Concepts

Balanced Chemical Equation

In every chemical reaction, it's essential to identify and represent the substances involved. A balanced chemical equation shows this, ensuring that the same number of each type of atom is present on both sides of the reaction. The equation for the reaction between sodium carbonate and silver nitrate is:\[ \text{Na}_2\text{CO}_3 (aq) + 2\text{AgNO}_3 (aq) \rightarrow \text{Ag}_2\text{CO}_3 (s) + 2\text{NaNO}_3 (aq) \]This equation demonstrates the relationship between reactants and products. Each side of the equation has the same number of atoms for each element. In this case, 1 mole of sodium carbonate combines with 2 moles of silver nitrate to yield 1 mole of silver carbonate and 2 moles of sodium nitrate. This balance is crucial for performing any stoichiometric calculations needed for predicting product formation.

Molar Mass Calculation

Understanding molar mass is foundational in chemistry, as it allows us to convert between grams and moles, the basic unit for chemical reactions. To find the molar mass, you sum the atomic masses of all atoms in a compound. For instance:

• Sodium Carbonate (Na₂CO₃): Calculated by \(2\times 22.99 + 12.01 + 3\times 16.00 = 105.99\ \text{g/mol}\).


• Silver Nitrate (AgNO₃): \(107.87 + 14.01 + 3\times 16.00 = 169.87\ \text{g/mol}\).


• Silver Carbonate (Ag₂CO₃): \(2\times 107.87 + 12.01 + 3\times 16.00 = 275.75\ \text{g/mol}\).


• Sodium Nitrate (NaNO₃): \(22.99 + 14.01 + 3\times 16.00 = 84.99\ \text{g/mol}\).

These calculations help in determining how much of each substance is present in a given sample, which is necessary for stoichiometry.

Stoichiometry

Stoichiometry is a key concept that involves the calculation of reactants and products in chemical reactions. It relies heavily on the balanced chemical equation and molar masses to predict how much of each element is involved. To determine the amount of product formed or reactant used, we apply mole ratios derived from the balanced equation. In this reaction:

• We begin by converting given masses to moles using their molar masses.


• Then, use stoichiometric coefficients (from the balanced equation) to find how reactants convert into products.


• Identify the limiting reactant—the one completely used up, determining maximum product formation.

Through stoichiometry, we find that silver nitrate limits the reaction, meaning it's the limiting reactant. This helps predict the amount of each product formed.

Product Formation

Product formation in any chemical reaction is governed by the amounts of reactants and the stoichiometry of the reaction. In this example, we calculated product formation based on the limiting reactant, silver nitrate. By knowing the moles of limiting reactant, we apply stoichiometry to predict how much of each product forms:

• Silver Carbonate: With a 1:2 ratio, \(\frac{0.0294}{2} = 0.0147\ \text{mol of Ag}_{2}\text{CO}_{3}\) formed, equaling \(0.0147\times 275.75\ \text{g/mol} \approx 4.05\ \text{g}\).


• Sodium Nitrate: Directly equivalent in ratio at 0.0294 mol, yielding \(0.0294 \times 84.99\ \text{g/mol} \approx 2.50 \ \text{g}\).

Understanding product formation is central for calculating yields and confirms that the reaction follows the principles of conservation of mass. This ensures no atoms are lost, only rearranged, allowing precise determination of products in chemical processes.