Problem 145 - Chemistry The Molecular Science

Question

When solutions of silver nitrate and sodium carbonate are mixed, solid silver carbonate is formed and sodium nitrate remains in solution. If a solution containing \(12.43 \mathrm{~g}\) sodium carbonate is mixed with a solution containing \(8.37 \mathrm{~g}\) silver nitrate, calculate the mass of each of the four substances present after the reaction is complete.

Step-by-Step Solution

Answer

6.80 g Ag2CO3, 4.19 g NaNO3, 9.82 g Na2CO3, 0 g AgNO3 remain.

Step 1: Write the Balanced Chemical Equation

Start by writing the balanced chemical equation for the reaction:\[ 2 \text{AgNO}_3 + \text{Na}_2\text{CO}_3 \rightarrow \text{Ag}_2\text{CO}_3 (s) + 2 \text{NaNO}_3 \]This equation tells us that two moles of silver nitrate react with one mole of sodium carbonate to produce one mole of silver carbonate (a solid) and two moles of sodium nitrate.

Step 2: Calculate the Molar Masses

Next, calculate the molar masses of each compound involved in the reaction:- \( \text{AgNO}_3 \): \( 107.87 \text{ g/mol (Ag)} + 14.01 \text{ g/mol (N)} + 3 \times 16.00 \text{ g/mol (O)} = 169.87 \text{ g/mol} \)- \( \text{Na}_2\text{CO}_3 \): \( 2 \times 22.99 \text{ g/mol (Na)} + 12.01 \text{ g/mol (C)} + 3 \times 16.00 \text{ g/mol (O)} = 105.99 \text{ g/mol} \)- \( \text{Ag}_2\text{CO}_3 \): \( 2 \times 107.87 \text{ g/mol (Ag)} + 12.01 \text{ g/mol (C)} + 3 \times 16.00 \text{ g/mol (O)} = 275.75 \text{ g/mol} \)\,- \( \text{NaNO}_3 \): \( 22.99 \text{ g/mol (Na)} + 14.01 \text{ g/mol (N)} + 3 \times 16.00 \text{ g/mol (O)} = 84.99 \text{ g/mol} \)\.

Step 3: Calculate Moles of Reactants

Determine the number of moles of each reactant:- Moles of \( \text{AgNO}_3 \): \( \frac{8.37 \text{ g}}{169.87 \text{ g/mol}} = 0.0493 \text{ mol} \).- Moles of \( \text{Na}_2\text{CO}_3 \): \( \frac{12.43 \text{ g}}{105.99 \text{ g/mol}} = 0.1173 \text{ mol} \).

Step 4: Determine Limiting Reactant

Compare the mole ratio according to the balanced equation to find the limiting reactant:- According to the equation, 2 moles of \( \text{AgNO}_3 \) are required per mole of \( \text{Na}_2\text{CO}_3 \).- As per our calculations, we need \( 0.1173 \times 2 = 0.2346 \text{ mol} \) of \( \text{AgNO}_3 \) to react with all of the \( \text{Na}_2\text{CO}_3 \).- We only have 0.0493 mol \( \text{AgNO}_3 \), which is less than 0.2346 mol required, so \( \text{AgNO}_3 \) is the limiting reactant.

Step 5: Calculate Moles of Products Formed

Calculate the moles of products formed, based on the moles of the limiting reactant (\( \text{AgNO}_3 \)):- Moles of \( \text{Ag}_2\text{CO}_3 \) formed: \( \frac{0.0493}{2} = 0.02465 \text{ mol} \), using the 2:1 mole ratio.- Moles of \( \text{NaNO}_3 \) formed: \( 0.0493 \text{ mol} \), as it follows a 1:1 ratio with \( \text{AgNO}_3 \).

Step 6: Calculate Mass of Products and Excess Reactant

Convert moles of products and the excess reactant to mass:- Mass of \( \text{Ag}_2\text{CO}_3 \) formed: \( 0.02465 \text{ mol} \times 275.75 \text{ g/mol} = 6.80 \text{ g} \).- Mass of \( \text{NaNO}_3 \) formed: \( 0.0493 \text{ mol} \times 84.99 \text{ g/mol} = 4.19 \text{ g} \).- Remaining moles of \( \text{Na}_2\text{CO}_3 \): \( 0.1173 - \frac{0.0493}{2} = 0.09265 \text{ mol} \).- Mass of remaining \( \text{Na}_2\text{CO}_3 \): \( 0.09265 \text{ mol} \times 105.99 \text{ g/mol} = 9.82 \text{ g} \).

Step 7: Summarize Results

The final masses of each substance after the reaction are:- \( \text{Ag}_2\text{CO}_3 \): 6.80 g (solid precipitate)- \( \text{NaNO}_3 \): 4.19 g (in solution)- Remaining \( \text{Na}_2\text{CO}_3 \): 9.82 g (unreacted, in solution)- \( \text{AgNO}_3 \): Completely reacted, so 0 g remains.

Key Concepts

Limiting Reactant

In chemical reactions, the concept of a limiting reactant is crucial because it determines how much product can be formed. The limiting reactant is the reactant that is completely consumed first, thus limiting the amount of product that can be formed.
For example, in the reaction between silver nitrate and sodium carbonate,

• We calculated that silver nitrate ( \( \text{AgNO}_3 \)) was the limiting reactant because we had only 0.0493 mol of it.
• Meanwhile, sodium carbonate (\( \text{Na}_2\text{CO}_3 \)) had a higher molar amount (0.1173 mol).
• According to the balanced chemical equation, we needed twice as much \( \text{AgNO}_3 \) to fully react with \( \text{Na}_2\text{CO}_3 \), but we had an insufficient amount, making it the limiting reactant.

Determining the limiting reactant is essential for predicting the quantities of products that will result from a reaction.

Stoichiometry

Stoichiometry refers to the calculation of reactants and products in chemical reactions. It is a way to relate the quantitative aspects of chemical reactions based on the ratios given in the balanced equation.
This involves the conversion from grams to moles, and the use of these ratios to determine how many moles of a product can be formed from given amounts of reactants. In our example:

• Start with converting the given masses of silver nitrate and sodium carbonate to moles using their molar masses.
• From the balanced equation \( 2\text{AgNO}_3 + \text{Na}_2\text{CO}_3 \rightarrow \text{Ag}_2\text{CO}_3 + 2 \text{NaNO}_3 \), we see that 2 moles of \( \text{AgNO}_3 \) react with 1 mole of \( \text{Na}_2\text{CO}_3 \).
• Use the concept of stoichiometry to calculate the theoretical yield of each product, which in this case were \( \text{Ag}_2\text{CO}_3 \) and \( \text{NaNO}_3 \).

Stoichiometry is the backbone of predicting product amounts, understanding reaction yields, and scaling up industrial chemical processes.

Balanced Chemical Equation

A balanced chemical equation is fundamental to understanding chemical reactions. It ensures that there are equal numbers of each type of atom on both sides of the equation, representing the conservation of mass.
In our reaction between silver nitrate and sodium carbonate:

• The balanced equation \[ 2 \text{AgNO}_3 + \text{Na}_2\text{CO}_3 \rightarrow \text{Ag}_2\text{CO}_3 (s) + 2 \text{NaNO}_3 \] shows that two molecules of silver nitrate react with one molecule of sodium carbonate, resulting in one molecule of silver carbonate and two molecules of sodium nitrate.
• This balance is crucial for accurate stoichiometric calculations, where the coefficients in the equation are used to determine the required and produced amounts of substances.
• The balanced equation provides a map of the reactants transforming into products, ensuring that mass is conserved throughout the reaction.

Understanding how to balance equations is key to predicting the outcome of reactions and ensuring no atoms are lost or gained in the process.