Problem 57 - Chemistry An Atoms First Approach

Question

What mass of \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) is required to precipitate all of the silver ions from \(75.0 \mathrm{mL}\) of a \(0.100-M\) solution of \(\mathrm{AgNO}_{3} ?\)

Step-by-Step Solution

Answer

The mass of \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) required to precipitate all silver ions from \(75.0 \mathrm{mL}\) of a \(0.100 \mathrm{M}\) solution of \(\mathrm{AgNO}_{3}\) can be calculated in the following manner: \(Moles\ of\ AgNO\text{_3} = \left(\frac{75.0}{1000}\right) \times 0.100\) \(Moles\ of\ Na\text{_2}CrO\text{_4} = \frac{1}{2} \times Moles\ of\ AgNO\text{_3}\) \(Mass\ of\ Na\text{_2}CrO\text{_4} = Moles\ of\ Na\text{_2}CrO\text{_4} \times (2 \times 22.99 + 51.996 + 4 \times 16)\)

Step 1: Calculate moles of Ag+ ions

First, we need to find the moles of the silver ions in the given solution. We are given that the solution's volume is \(75.0 \mathrm{mL}\), and its concentration is \(0.100 \mathrm{M}\). To calculate the moles, multiply the volume of the solution (in liters) by its concentration: Moles of AgNO3 = Volume (in L) × Concentration \(=\left( \frac{75.0}{1000} \right) \times 0.100\)

Step 2: Determine stoichiometric ratio

Next, we determine the stoichiometric ratio between AgNO3 and Na2CrO4 from the balanced chemical equation: 2AgNO3 (aq) + Na2CrO4 (aq) → Ag2CrO4 (s) + 2NaNO3 (aq) The stoichiometric ratio between AgNO3 and Na2CrO4 is 2:1.

Step 3: Calculate moles of Na2CrO4 needed

Given the 2:1 stoichiometric ratio, we require half the number of moles of Na2CrO4 as there were moles of AgNO3: Moles of Na2CrO4 \(= \frac{1}{2}\) × Moles of AgNO3

Step 4: Calculate mass of Na2CrO4 required

Now, we need to find the mass of Na2CrO4 required. To do this, multiply the moles of Na2CrO4 calculated above by the molar mass of Na2CrO4: Mass of Na2CrO4 = Moles of Na2CrO4 × Molar mass of Na2CrO4 Use the periodic table to find the molar mass of Na2CrO4: \(Molar\ mass\ of\ \mathrm{Na}_{2} \mathrm{CrO}_{4} = 2 \times (Na) + 1 \times (Cr) + 4 \times (O) = 2 \times 22.99 + 51.996 + 4 \times 16\) Now combine all calculations, and find the mass of Na2CrO4 required to precipitate all silver ions: Moles of AgNO3 \(= \left(\frac{75.0}{1000}\right) \times 0.100\) Moles of Na2CrO4 \(= \frac{1}{2} \times \text{Moles of AgNO}\text{_3}\) Mass of Na2CrO4 \(= \text{Moles of Na}\text{_2}\text{CrO}\text{_4} \times \text{Molar mass of Na}\text{_2}\text{CrO}\text{_4}\)

Key Concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It is a key part of understanding chemical equations and their implications. When solving stoichiometry problems, we use the coefficients of each substance in a balanced chemical equation to determine the proportion of reactants and products. The stoichiometric coefficients represent the molar ratio in which substances react and form products.

For instance, in a precipitation reaction where silver nitrate (\text{AgNO}_3) reacts with sodium chromate (\text{Na}_2\text{CrO}_4), the balanced chemical equation reads: \[2\text{AgNO}_3(aq) + \text{Na}_2\text{CrO}_4(aq) \rightarrow \text{Ag}_2\text{CrO}_4(s) + 2\text{NaNO}_3(aq)\].Here, the stoichiometric ratio of \text{AgNO}_3 to \text{Na}_2\text{CrO}_4 is 2 to 1. This ratio implies that two moles of \text{AgNO}_3 react with one mole of \text{Na}_2\text{CrO}_4, which is crucial when calculating the amount of \text{Na}_2\text{CrO}_4 needed to precipitate all the silver ions present.

Molar Mass Calculation

The molar mass of a substance is the weight of one mole of that substance, usually expressed in grams per mole (g/mol). It is calculated by adding up the atomic masses of all the elements present in one molecule of the compound. Atomic masses can be found on the periodic table, where they are listed in atomic mass units (amu) approximated to two decimal places.

Using sodium chromate (\text{Na}_2\text{CrO}_4) as an example, its molar mass is determined by adding up the molar masses of sodium (Na), chromium (Cr), and oxygen (O). For \text{Na}_2\text{CrO}_4, this calculation would look like: \[Molar\text{ mass of }\text{Na}_2\text{CrO}_4 = 2 \times (Molar\text{ mass of Na}) + (Molar\text{ mass of Cr}) + 4 \times (Molar\text{ mass of O})\].Given the atomic masses of Na (22.99 amu), Cr (51.996 amu), and O (16 amu), the molar mass of sodium chromate is then \[2 \times 22.99 + 51.996 + 4 \times 16 \] g/mol. Accurate calculation of molar mass is essential for converting between amounts of a substance in moles and its corresponding mass in grams.

Chemical Equations

Chemical equations are symbolic representations of chemical reactions, showing the reactants on the left and the products on the right. They are balanced according to the law of conservation of mass, meaning that the number of atoms for each element must be the same on both sides of the equation. Writing and balancing chemical equations are critical skills for solving stoichiometry problems.

In the reaction between silver nitrate and sodium chromate, we represent it as: \[2\text{AgNO}_3(aq) + \text{Na}_2\text{CrO}_4(aq) \rightarrow \text{Ag}_2\text{CrO}_4(s) + 2\text{NaNO}_3(aq)\].This equation indicates that two moles of silver nitrate react with one mole of sodium chromate to produce silver chromate and sodium nitrate. Understanding the balanced chemical equation helps in predicting the outcome of a reaction, such as the formation of a precipitate in this case, which is crucial when the goal is to determine the mass of \text{Na}_2\text{CrO}_4 required to fully precipitate silver ions from a solution.