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 between sodium carbonate (Na2CO3) and silver nitrate (AgNO3) is complete, the following masses of substances are present: \(1.93\,g\) of Na2CO3 remaining, \(0\,g\) of AgNO3 remaining, \(1.57\,g\) of Ag2CO3 formed, and \(5.00\,g\) of NaNO3 formed.

Step-by-step solution

Write the balanced chemical equation

The balanced chemical equation for the reaction between sodium carbonate (Na2CO3) and silver nitrate (AgNO3) to form silver carbonate (Ag2CO3) and sodium nitrate (NaNO3) is: \(Na_2CO_3 + 2AgNO_3 \rightarrow Ag_2CO_3 + 2NaNO_3\)

Calculate the moles of each reactant

To determine which reactant is the limiting reactant, the moles of sodium carbonate (Na2CO3) and silver nitrate (AgNO3) are needed. First, we can find the molar masses of Na2CO3 and AgNO3. \(Molar\, mass_{Na_2CO_3} = 2\times23 + 12 + 3\times16 = 106\, g/mol\) \(Molar\, mass_{AgNO_3} = 108 + 14 + 3\times16 = 170\, g/mol\) Now we can calculate the moles of Na2CO3 and AgNO3 using the given masses: \(moles_{Na_2CO_3} = \frac{3.50\, g}{106\, g/mol} = 0.03302\, mol\) \(moles_{AgNO_3} = \frac{5.00\, g}{170\, g/mol} = 0.02941\, mol\)

Determine the limiting reactant

Using the stoichiometry of the balanced equation, examine the mole ratio of Na2CO3 and AgNO3: \(\frac{moles_{Na_2CO_3}}{moles_{AgNO_3}} = \frac{0.03302\, mol}{0.02941\, mol} = 1.12\) According to the stoichiometry of the balanced equation, the ratio of Na2CO3 to AgNO3 should be \(1:2\). Since we observe a higher ratio here, it shows that we have more sodium carbonate than needed for the reaction. Therefore, silver nitrate (AgNO3) is the limiting reactant.

Calculate the moles of products formed

Based on the limiting reactant, AgNO3, we can use stoichiometry to calculate the moles of products formed after the reaction. \(moles_{Ag_2CO_3} = \frac{1}{2} \times moles_{AgNO_3} = \frac{1}{2} \times 0.02941\, mol = 0.01471\, mol\) \(moles_{NaNO_3} = 2 \times moles_{AgNO_3} = 2 \times 0.02941\, mol = 0.05882\, mol\)

Calculate the moles of reactants remaining after the reaction

We can also calculate how much sodium carbonate (Na2CO3) remains after the reaction: \(moles_{Na_2CO_3\, remaining} = moles_{Na_2CO_3\, initial} - 0.5 \times moles_{AgNO_3} = 0.03302\, mol - 0.5 \times 0.02941\, mol = 0.01816\, mol\) Since AgNO3 is the limiting reactant, there will be no AgNO3 left after the reaction.

Convert the moles of substances to grams

Convert moles of all substances to grams after the reaction is complete by multiplying the moles by their respective molar masses: \(mass_{Na_2CO_3\, remaining} = 0.01816\, mol \times 106\, g/mol = 1.93\, g\) \(mass_{AgNO_3\, remaining} = 0\, g \) \(mass_{Ag_2CO_3\, formed} = 0.01471\, mol \times (108\times2 + 12 + 3\times16)\, g/mol = 1.57\, g\) \(mass_{NaNO_3\, formed} = 0.05882\, mol \times 85\, g/mol = 5.00\, g\)

Write the final answer

After the reaction between sodium carbonate (Na2CO3) and silver nitrate (AgNO3), the masses of the substances present are: - Sodium carbonate (Na2CO3) remaining: \(1.93\, g\) - Silver nitrate (AgNO3) remaining: \(0\, g\) - Silver carbonate (Ag2CO3) formed: \(1.57\, g\) - Sodium nitrate (NaNO3) formed: \(5.00\, g\)

Key concepts

Limiting Reactant

Understanding the limiting reactant concept is fundamental in performing stoichiometric calculations. It's akin to running out of eggs when baking a cake: no matter how much of the other ingredients you have, you can't make more cakes without more eggs.

In chemistry, the limiting reactant is the substance that is totally consumed first during a chemical reaction, thus determining the amount of product produced. This is because chemical reactions occur according to defined mole ratios of reactants and products as described by the balanced chemical equation. If one of the reactants is used up before the others, the reaction stops, even if other reactants are still available.

In the example provided, after calculating the molar amounts of sodium carbonate and silver nitrate, a comparison of their ratios to the stoichiometric ratios shows that silver nitrate is the limiting reactant. This means that all of the silver nitrate will be consumed in the reaction, while some sodium carbonate will be left unreacted.

Stoichiometric Calculations

Stoichiometric calculations are the mathematical methods used to determine the quantities of reactants and products involved in chemical reactions. These calculations are based on the law of conservation of mass, which states that matter is neither created nor destroyed in a chemical reaction.

To perform these calculations, you must start with a balanced chemical equation, which gives the ratio in which reactants combine and products form. In the provided exercise, we see how the molar mass of each substance is used in combination with the given masses to calculate the moles of reactants. Then, using the mole ratio from the balanced equation, we can find out which reactant will limit the reaction and how much product can be formed.

Following the step by step solution, by knowing the limiting reactant (silver nitrate in this case), we can calculate the moles of all substances after the reaction and subsequently convert these moles back into grams. This process underpins most of quantitative chemistry and is essential for predicting the outcomes of reactions.

Molar Mass

The molar mass is a critical factor in stoichiometric calculations as it links the mass of a substance with the amount in moles, thereby bridging the macroscopic and microscopic worlds of chemistry. It is defined as the mass of one mole of a given substance, typically expressed in grams per mole (g/mol).

To find the molar mass, you sum the masses of all the atoms in one molecule of the substance. For example, in the exercise, you calculated the molar mass of sodium carbonate (Na2CO3) and silver nitrate (AgNO3) by adding the atomic masses of their constituent atoms. Once you have the molar mass, you can convert grams to moles and vice versa. This is essential when translating the theoretical stoichiometry from a balanced equation into actual masses that can be measured in a lab.