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?

The balanced chemical equation for the reaction between sodium carbonate (Na2CO3) and silver nitrate (AgNO3) is: Na2CO3(aq) + 2 AgNO3(aq) -> 2 Ag2CO3(s) + 2 NaNO3(aq)

Molar mass of sodium carbonate (Na2CO3) = (2*22.99) + (12.01) + (3*16.00) = 105.98 g/mol Molar mass of silver nitrate (AgNO3) = (107.87) + (14.01) + (3*16.00) = 169.88 g/mol Molar mass of silver carbonate (Ag2CO3) = (2*107.87) + (12.01) + (3*16.00) = 275.75 g/mol Molar mass of sodium nitrate (NaNO3) = (22.99) + (14.01) + (3*16.00) = 85.00 g/mol

Given masses of reactants are: Na2CO3 = 3.50 g AgNO3 = 5.00 g Moles of sodium carbonate (Na2CO3) = 3.50 g / 105.98 g/mol ≈ 0.0330 mol Moles of silver nitrate (AgNO3) = 5.00 g / 169.88 g/mol ≈ 0.0294 mol

From the balanced equation: 1 mol of Na2CO3 requires 2 mol of AgNO3 We have 0.0330 mol of Na2CO3 and 0.0294 mol of AgNO3. Let's check if there is enough AgNO3 to react with Na2CO3: 0.0330 mol of Na2CO3 * 2 mol of AgNO3 = 0.0660 mol of required AgNO3 Since we have only 0.0294 mol of AgNO3, it is the limiting reactant, and the reaction will stop when it is completely consumed.

As the reaction proceeds, all 0.0294 mol of AgNO3 will be consumed. To find the amount of Na2CO3 consumed, we will use the stoichiometry from the balanced equation: 0.0294 mol of AgNO3 * (1 mol Na2CO3 / 2 mol AgNO3) = 0.0147 mol of Na2CO3 consumed So, after the reaction, the amount of Na2CO3 left = initial amount - consumed amount = 0.0330 - 0.0147 = 0.0183 mol Then, the mass of Na2CO3 left = 0.0183 mol * 105.98 g/mol ≈ 1.94 g Now let's find the mass of products formed: Silver carbonate (Ag2CO3) formed = 0.0294 mol of AgNO3 * (1 mol Ag2CO3 / 2 mol AgNO3) = 0.0147 mol of Ag2CO3 Mass of silver carbonate (Ag2CO3) formed = 0.0147 mol * 275.75 g/mol ≈ 4.06 g Sodium nitrate (NaNO3) formed = 0.0294 mol AgNO3 (1 mol NaNO3 / 1 mol AgNO3) = 0.0294 mol NaNO3 Mass of sodium nitrate (NaNO3) formed = 0.0294 mol * 85.00 g/mol ≈ 2.50 g Finally, after the reaction is complete, the amounts of each compound are: Mass of sodium carbonate (Na2CO3) left = 1.94 g Mass of silver nitrate (AgNO3) left = 0 g (all consumed) Mass of silver carbonate (Ag2CO3) formed = 4.06 g Mass of sodium nitrate (NaNO3) formed = 2.50 g