91Ó°ÊÓ

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

Using the molar masses of sodium carbonate (Na₂CO₃) and silver nitrate (AgNO₃), we can calculate the moles of each reactant: Moles of sodium carbonate = (3.50 g Na₂CO₃) / (105.99 g/mol Na₂CO₃) = 0.0330 mol Na₂CO₃ Moles of silver nitrate = (5.00 g AgNO₃) / (169.87 g/mol AgNO₃) = 0.0294 mol AgNO₃

To determine the limiting reactant, we can compare the mole ratio of sodium carbonate to silver nitrate in the balanced equation (1:2) with the moles of the reactants given: 0.0330 mol Na₂CO₃ / 1 = 0.0330 mol Na₂CO₃ 0.0294 mol AgNO₃ / 2 = 0.0147 mol AgNO₃ Since 0.0147 mol AgNO₃ < 0.0330 mol Na₂CO₃, silver nitrate (AgNO₃) is the limiting reactant.

Using the stoichiometry of the balanced chemical equation, we can determine the moles of silver carbonate and sodium nitrate produced: Moles of silver carbonate (Ag₂CO₃) = 0.0294 mol AgNO₃ × (1 mol Ag₂CO₃ / 2 mol AgNO₃) = 0.0147 mol Ag₂CO₃ Moles of sodium nitrate (NaNO₃) = 0.0294 mol AgNO₃ × (2 mol NaNO₃ / 2 mol AgNO₃) = 0.0294 mol NaNO₃

Finally, we can determine the mass of each compound present after the reaction is complete: Mass of sodium carbonate (Na₂CO₃) = 0.0330 mol Na₂CO₃ - (0.0147 mol Ag₂CO₃ × (1 mol Na₂CO₃ / 1 mol Ag₂CO₃)) = 0.0183 mol Na₂CO₃ × (105.99 g/mol Na₂CO₃) = 1.94 g Na₂CO₃ Mass of silver nitrate (AgNO₃) = 0 - AgNO₃ is fully consumed in the reaction Mass of silver carbonate (Ag₂CO₃) = 0.0147 mol Ag₂CO₃ × (275.74 g/mol Ag₂CO₃) = 4.05 g Ag₂CO₃ Mass of sodium nitrate (NaNO₃) = 0.0294 mol NaNO₃ × (85.00 g/mol NaNO₃) = 2.50 g NaNO₃ So, after the reaction is complete, there are 1.94 g of sodium carbonate, 0 g of silver nitrate, 4.05 g of silver carbonate, and 2.50 g of sodium nitrate present.