Question

Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A solution containing 4.50 g of sodium carbonate is mixed with one containing 3.25 g of silver nitrate. After the reaction is complete, the solutions are evaporated to dryness, leaving a mixture of salts. How many grams of each of the following compounds are present after the reaction is complete?

sodium carbonate?
silver nitrate?
silver carbonate?
sodium nitrate?

Answer 1

Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A solution containing $4.50\mathrm{g}$ of sodium carbonate is mixed with one containing $3.25\mathrm{g}$ of silver nitrate. After the reaction is complete, the solutions are evaporated to dryness, leaving a mixture of salts. How many grams of each of the following compounds are present after the reaction is complete?

- sodium carbonate?

- silver nitrate?

- silver carbonate?

- sodium nitrate?

Solution:

$$2 \mathrm{AgNO_3} + \mathrm{Na_2CO_3} \rightarrow \mathrm{Ag_2CO_3} + 2 \mathrm{NaNO_3}$$ $$\begin{array}{l} n_1 \left(\mathrm{Na_2CO_3}\right) = \frac{4.50 \mathrm{g}}{105.9 \mathrm{g/mol}} = 0.04249 \mathrm{mol} \\ n_1 \left(\mathrm{AgNO_3}\right) = \frac{3.25 \mathrm{g}}{169.87 \mathrm{g/mol}} = 0.01913 \mathrm{mol} \\ n_2 \left(\mathrm{Na_2CO_3}\right) = \frac{0.01913 \mathrm{mol}}{2} = 0.009565 \mathrm{mol} \\ n_3 \left(\mathrm{Na_2CO_3}\right) = 0.04249 \mathrm{mol} - 0.009565 \mathrm{mol} = 0.032925 \mathrm{mol} \\ n \left(\mathrm{Ag_2CO_3}\right) = 0.009565 \mathrm{mol} \\ n \left(\mathrm{NaNO_3}\right) = 0.01913 \mathrm{mol} \\ m \left(\mathrm{Na_2CO_3}\right) = n_3 \cdot M = 0.032925 \mathrm{mol} \cdot 105.9 \mathrm{g/mol} = 3.4867 \mathrm{g} \\ m \left(\mathrm{Ag_2CO_3}\right) = 0.009565 \mathrm{mol} \cdot 275.74 \mathrm{g/mol} = 2.6374 \mathrm{g} \\ m \left(\mathrm{NaNO_3}\right) = 0.01913 \mathrm{mol} \cdot 84.9 \mathrm{g/mol} = 1.6241 \mathrm{g} \\ \end{array}$$

Answer:

$$\begin{array}{l} m \left(\mathrm{AgNO_3}\right) = 0 \mathrm{g} \\ m \left(\mathrm{Na_2CO_3}\right) = 3.4867 \mathrm{g} \\ m \left(\mathrm{Ag_2CO_3}\right) = 2.6374 \mathrm{g} \\ m \left(\mathrm{NaNO_3}\right) = 1.6241 \mathrm{g} \\ \end{array}$$

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