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: \( Na_2CO_3(aq) + 2AgNO_3(aq) \rightarrow 2NaNO_3(aq) + Ag_2CO_3(s) \)

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) \(1 \cdot 22.99 + 1 \cdot 14.01 + 3 \cdot 16.00 = 85.00 \mathrm{g/mol}\)

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: \( \frac{5.00 \mathrm{~g}}{169.87 \frac{\mathrm{g}}{\mathrm{mol}}} = 0.0294 \mathrm{~mol} \)

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: \( 0.0294 \mathrm{~mol \, AgNO_3} \cdot \frac{1\, \mathrm{mol \, Ag_2CO_3}}{2\, \mathrm{mol \, AgNO_3}} = 0.0147 \mathrm{~mol \, Ag_2CO_3} \)

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.

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