Problem 81 - Chemistry: The Central Science

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

After the reaction is complete, there will be 1.94 g of sodium carbonate, 0 g of silver nitrate (completely reacted), 2.03 g of silver carbonate, and 1.25 g of sodium nitrate.

Step 1: Write the balanced chemical equation

For the reaction between sodium carbonate (Na2CO3) and silver nitrate (AgNO3) to form silver carbonate (Ag2CO3) and sodium nitrate (NaNO3), the balanced chemical equation is: \[Na_2CO_3 + 2AgNO_3 \rightarrow Ag_2CO_3 + 2NaNO_3\]

Step 2: Convert the given masses to moles

Using the molar masses of sodium carbonate (105.99 g/mol) and silver nitrate (169.87 g/mol), we can convert their masses to moles: - Moles of Na2CO3: \(\frac{3.50 \, g}{105.99 \, g/mol}=0.0330 \, mol\) - Moles of AgNO3: \(\frac{5.00 \, g}{169.87 \, g/mol}=0.0294 \, mol\)

Step 3: Determine the limiting reactant

Compare the mole ratio given by the balanced equation with the mole ratio from the given mass information. The balanced equation tells us that 1 mol of Na2CO3 reacts with 2 mol of AgNO3. Now, we divide the moles of each reactant by their respective coefficients in the balanced equation: - For Na2CO3: \(0.0330 \, mol \div 1 = 0.0330\) - For AgNO3: \(0.0294 \, mol \div 2 = 0.0147\) Since the minimum ratio is 0.0147 for AgNO3, it is the limiting reactant.

Step 4: Calculate the moles of products formed

Using the stoichiometry of the balanced equation, calculate the moles of silver carbonate (Ag2CO3) and sodium nitrate (NaNO3) formed: - Moles of Ag2CO3: \(0.0147 \, mol \times 1 \, mol \, Ag_2CO_3 / 2 \, mol \, AgNO_3 = 0.00735 \, mol\) - Moles of NaNO3: \(0.0147 \, mol \times 2 \, mol \, NaNO_3 / 2 \, mol \, AgNO_3 = 0.0147 \, mol\)

Step 5: Convert moles of products to grams

Using the molar masses of silver carbonate (275.75 g/mol) and sodium nitrate (85.00 g/mol), we can convert the moles of products to grams: - Mass of Ag2CO3: \(0.00735 \, mol \times 275.75 \, g/mol = 2.03 \, g\) - Mass of NaNO3: \(0.0147 \, mol \times 85.00 \, g/mol = 1.25 \, g\) Now we need to find out the remaining mass of sodium carbonate after the reaction: - Mass of remaining Na2CO3: \(3.5 \, g - (0.0330 \, mol - 0.0147 \, mol) \times 105.99 \, g/mol = 1.94 \, g\) Finally, we have the masses of all compounds after the reaction is complete: - Sodium carbonate: 1.94 g - Silver nitrate: 0 g (completely reacted) - Silver carbonate: 2.03 g - Sodium nitrate: 1.25 g