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?

Short answer

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.

Step-by-step solution

Write the balanced chemical equation

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

Calculate the moles of sodium carbonate and silver nitrate

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

Determine the limiting reactant

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.

Calculate the moles of the products produced by the reaction

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

Convert the moles of each compound to grams

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.